If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4, then the sum of its first twelve terms is

### Related Formula ntext-th term of a GP: T_n = a r^n-1$$n\text{-th term of a GP}: T_n = a r^{n-1}$$ ### Core Logic Let the terms of the GP be a, ar, ar^2, ar^3, dots$a, ar, ar^2, ar^3, \dots$ Given: a_2 cdot a_3 cdot a_4 = 64$a_2 \cdot a_3 \cdot a_4 = 64$ (ar) cdot (ar^2) cdot (ar^3) = 64$$(ar) \cdot (ar^2) \cdot (ar^3) = 64$$ a^3 r^6 = 64 Rightarrow (ar^2)^3 = 64 Rightarrow ar^2 = 4$$a^3 r^6 = 64 \Rightarrow (ar^2)^3 = 64 \Rightarrow ar^2 = 4$$ Since it is a GP of increasing positive terms, r > 1$r > 1$ and a > 0$a > 0$. ### Step 1: Utilize the sum condition Given: a_1 + a_3 + a_5 = frac8137$a_1 + a_3 + a_5 = \frac{813}{7}$ a + ar^2 + ar^4 = frac8137$$a + ar^2 + ar^4 = \frac{813}{7}$$ Extract a$a$ from ar^2 = 4 Rightarrow a = frac4r^2$ar^2 = 4 \Rightarrow a = \frac{4}{r^2}$. Substitute this in: frac4r^2 + 4 + 4r^2 = frac8137$$\frac{4}{r^2} + 4 + 4r^2 = \frac{813}{7}$$ 4left(frac1r^2 + 1 + r^2right) = frac8137$$4\left(\frac{1}{r^2} + 1 + r^2\right) = \frac{813}{7}$$ Let r^2 = t$r^2 = t$: 4left(frac1t + 1 + tright) = frac8137$$4\left(\frac{1}{t} + 1 + t\right) = \frac{813}{7}$$ Wait, there is a much faster method by just scaling the required expression. ### Step 2: Calculate the required expression We need a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$a_3 + a_5 + a_7 = ar^2 + ar^4 + ar^6$. Notice that ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$ar^2 + ar^4 + ar^6 = r^2 (a + ar^2 + ar^4)$. So, required sum = r^2 left(frac8137right)$r^2 \left(\frac{813}{7}\right)$. To find r^2$r^2$, we solve the quadratic in t = r^2$t = r^2$: 4left(fract^2 + t + 1tright) = frac8137$$4\left(\frac{t^2 + t + 1}{t}\right) = \frac{813}{7}$$ 28t^2 + 28t + 28 = 813t$$28t^2 + 28t + 28 = 813t$$ 28t^2 - 785t + 28 = 0$$28t^2 - 785t + 28 = 0$$ The roots are t = 28$t = 28$ and t = frac128$t = \frac{1}{28}$. Since the GP is increasing, r > 1 Rightarrow r^2 = 28$r > 1 \Rightarrow r^2 = 28$. ### Step 3: Final evaluation Alternatively, expand directly: ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ar^2 (1 + r^2 + r^4) = 4 (1 + 28 + (28)^2) = 4(1 + 28 + 784) = 4(813) = 3252.$$ (Note: r^2 times frac8137 = 28 times frac8137 = 4 times 813 = 3252$r^2 \times \frac{813}{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252$) ### Pattern Recognition In GP questions demanding a sum shifted by a fixed index (like a_1+a_3+a_5$a_1+a_3+a_5$ to a_3+a_5+a_7$a_3+a_5+a_7$), immediately look to factor out the common ratio multiplier r^k$r^k$. Here it's a simple scaling by r^2$r^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series Class 10 Maths: Quadratic Equations

### Related Formula Partial fraction decomposition for telescopic summing: frac2n^2 - (n+1)^2 + 1 dotsdots text structures directly cancel in series expansions.$$\frac{2n^2 - (n+1)^2 + 1 \dots}{\dots} \text{ structures directly cancel in series expansions.}$$ ### Core Logic Given recurrence: a_n+1 - frac12a_n = fracn^2 - 2n - 1n^2(n+1)^2$$a_{n+1} - \frac{1}{2}a_n = \frac{n^2 - 2n - 1}{n^2(n+1)^2}$$ Rewrite the numerator to split the fraction: n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$n^2 - 2n - 1 = 2n^2 - (n^2 + 2n + 1) = 2n^2 - (n+1)^2$$ a_n+1 - frac12a_n = frac2n^2 - (n+1)^2n^2(n+1)^2 = frac2(n+1)^2 - frac1n^2$$a_{n+1} - \frac{1}{2}a_n = \frac{2n^2 - (n+1)^2}{n^2(n+1)^2} = \frac{2}{(n+1)^2} - \frac{1}{n^2}$$ ### Step 1: Telescope generation Multiply both sides by appropriate powers of 2 to create a cancelling chain: For n=1$n=1$: a_2 - frac12a_1 = frac22^2 - frac11^2$a_2 - \frac{1}{2}a_1 = \frac{2}{2^2} - \frac{1}{1^2}$ For n=2$n=2$: multiply by 2 Rightarrow 2left[a_3 - frac12a_2 = frac23^2 - frac12^2right] Rightarrow 2a_3 - a_2 = frac2 times 23^2 - frac22^2$\Rightarrow 2\left[a_3 - \frac{1}{2}a_2 = \frac{2}{3^2} - \frac{1}{2^2}\right] \Rightarrow 2a_3 - a_2 = \frac{2 \times 2}{3^2} - \frac{2}{2^2}$ Wait, let's look at a cleaner telescopic scaling: a_n+1 - frac2(n+1)^2 = frac12 left(a_n - frac2n^2right)$a_{n+1} - \frac{2}{(n+1)^2} = \frac{1}{2} \left(a_n - \frac{2}{n^2}\right)$. ### Step 2: Identify Geometric Progression Let V_n = a_n - frac2n^2$V_n = a_n - \frac{2}{n^2}$. The recurrence gives V_n+1 = frac12 V_n$V_{n+1} = \frac{1}{2} V_n$. This proves V_n$V_n$ is a geometric progression with common ratio r = 1/2$r = 1/2$. First term V_1 = a_1 - frac21^2 = 1 - 2 = -1$V_1 = a_1 - \frac{2}{1^2} = 1 - 2 = -1$. ### Step 3: Infinite Summation We need left| sum_n=1^infty left( a_n - frac2n^2 right) right| = left| sum_n=1^infty V_n right|$\left| \sum_{n=1}^{\infty} \left( a_n - \frac{2}{n^2} \right) \right| = \left| \sum_{n=1}^{\infty} V_n \right|$. Since V_n$V_n$ is an infinite GP: S_infty = fracV_11 - r = frac-11 - 1/2 = frac-11/2 = -2$$S_{\infty} = \frac{V_1}{1 - r} = \frac{-1}{1 - 1/2} = \frac{-1}{1/2} = -2$$ Taking absolute value: |-2| = 2$|-2| = 2$ ### Pattern Recognition When dealing with rational fraction recurrences A_n+1 - k A_n = f(n) - k f(n-1)$A_{n+1} - k A_n = f(n) - k f(n-1)$, immediately substitute V_n = A_n - f(n)$V_n = A_n - f(n)$. This substitution instantly isolates a classical Geometric Progression. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequences and Series

### Related Formula textSum of G.P. S_n = fraca(r^n - 1)r - 1 text (for r > 1)$$\text{Sum of G.P. } S_n = \frac{a(r^n - 1)}{r - 1} \text{ (for } r > 1)$$ ### Core Logic The given sequence is a G.P. with ratio frac1sqrt2$\frac{1}{\sqrt{2}}$. fraca_2/2a_1 = frac1sqrt2 implies a_2 = a_1 sqrt2$$\frac{a_2/2}{a_1} = \frac{1}{\sqrt{2}} \implies a_2 = a_1 \sqrt{2}$$ fraca_3/2^2a_2/2 = frac1sqrt2 implies fraca_32 a_2 = frac1sqrt2 implies a_3 = a_2 sqrt2 = a_1 (sqrt2)^2$$\frac{a_3/2^2}{a_2/2} = \frac{1}{\sqrt{2}} \implies \frac{a_3}{2 a_2} = \frac{1}{\sqrt{2}} \implies a_3 = a_2 \sqrt{2} = a_1 (\sqrt{2})^2$$ Thus, a_1, a_2, a_3, dots, a_10$a_1, a_2, a_3, \dots, a_{10}$ forms a standard G.P. with first term a_1$a_1$ and common ratio R = sqrt2$R = \sqrt{2}$. ### Step 1: Calculate the Sum Sum of this new sequence is S_10 = 62$S_{10} = 62$. S_10 = fraca_1 left( (sqrt2)^10 - 1 right)sqrt2 - 1 = 62$$S_{10} = \frac{a_1 \left( (\sqrt{2})^{10} - 1 \right)}{\sqrt{2} - 1} = 62$$ Since (sqrt2)^10 = 2^5 = 32$(\sqrt{2})^{10} = 2^5 = 32$: 62 = fraca_1 (32 - 1)sqrt2 - 1$$62 = \frac{a_1 (32 - 1)}{\sqrt{2} - 1}$$ 62 = frac31 a_1sqrt2 - 1$$62 = \frac{31 a_1}{\sqrt{2} - 1}$$ 2 = fraca_1sqrt2 - 1$$2 = \frac{a_1}{\sqrt{2} - 1}$$ a_1 = 2(sqrt2 - 1)$$a_1 = 2(\sqrt{2} - 1)$$ ### Pattern Recognition If a sequence b_n = fraca_nk^n-1$b_n = \frac{a_n}{k^{n-1}}$ is a G.P. with ratio r$r$, then the base sequence a_n$a_n$ is inherently a G.P. with ratio R = kr$R = kr$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Sequence and Series

### Related Formula textSum of an A.P.: S_k = frack2 left( 2a + (k-1)d right)$$\text{Sum of an A.P.: } S_k = \frac{k}{2} \left( 2a + (k-1)d \right)$$ textGeneral term of an A.P.: a_k = a_1 + (k-1)d$$\text{General term of an A.P.: } a_k = a_1 + (k-1)d$$ ### Core Logic Let the A.P. have n$n$ terms (where n$n$ is even). The terms are divided into n/2$n/2$ odd-indexed terms and n/2$n/2$ even-indexed terms. ### Step 1: Set up the even and odd sums Sum of even terms: a_2 + a_4 + dots + a_n = 30 quad text--- (1)$$a_2 + a_4 + \dots + a_n = 30 \quad \text{--- (1)}$$ Sum of odd terms: a_1 + a_3 + dots + a_n-1 = 24 quad text--- (2)$$a_1 + a_3 + \dots + a_{n-1} = 24 \quad \text{--- (2)}$$ Subtracting equation (2) from (1): (a_2 - a_1) + (a_4 - a_3) + dots + (a_n - a_n-1) = 30 - 24 = 6$$(a_2 - a_1) + (a_4 - a_3) + \dots + (a_n - a_{n-1}) = 30 - 24 = 6$$ Since there are n/2$n/2$ such pairs, and the difference of adjacent terms is the common difference d$d$: fracn2 d = 6 implies n d = 12 quad text--- (3)$$\frac{n}{2} d = 6 \implies n d = 12 \quad \text{--- (3)}$$ ### Step 2: Solve for n and d We are given that the last term exceeds the first by frac212$\frac{21}{2}$: a_n - a_1 = (n-1)d = frac212$$a_n - a_1 = (n-1)d = \frac{21}{2}$$ n d - d = 10.5$n d - d = 10.5$ Substitute nd = 12$nd = 12$ from (3): 12 - d = 10.5 implies d = 1.5 = frac32$$12 - d = 10.5 \implies d = 1.5 = \frac{3}{2}$$ Using this in (3): n left(frac32right) = 12 implies n = 8$$n \left(\frac{3}{2}\right) = 12 \implies n = 8$$ ### Step 3: Solve for the first term The sum of the odd terms is: S_textodd = frac42 left[ 2a_1 + (4-1)(2d) right] = 24$$S_{\text{odd}} = \frac{4}{2} \left[ 2a_1 + (4-1)(2d) \right] = 24$$ 2 left[ 2a_1 + 3(3) right] = 24 implies 2a_1 + 9 = 12 implies a_1 = 1.5 = frac32$$2 \left[ 2a_1 + 3(3) \right] = 24 \implies 2a_1 + 9 = 12 \implies a_1 = 1.5 = \frac{3}{2}$$ Thus, the terms are: frac32, \, 3, \, frac92, \, 6, \, frac152, \, 9, \, frac212, \, 12$$\frac{3}{2}, \, 3, \, \frac{9}{2}, \, 6, \, \frac{15}{2}, \, 9, \, \frac{21}{2}, \, 12$$ The terms that are integers are 3, 6, 9, 12$3, 6, 9, 12$. The total number of integer terms is 4. ### Pattern Recognition Sum of even terms minus sum of odd terms in any A.P. with an even number of terms n$n$ is always equal to fracn2 d$\frac{n}{2} d$. This is an extremely useful relation to remember. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Sequences and Series