JEE Main 2024 9 April Evening Shift
Exam JEE Main
Year 2024
Date 9 April
Shift Evening Shift
Subject Mathematics
Chapter Sequences and Series JEE Main 2024 · 9 April · Evening Shift · Mathematics · Sequences and Series
Let a, a r, a r^2 , be an infinite G.P. If ∑_ n=0 ^ ∞ a r^n=57 and...
Let
a , a r , a r 2 a, a r, a r^2 a , a r , a r 2 ,
\qquad be an infinite G.P. If
∑ n = 0 ∞ a r n = 57 \sum_{n=0}^{\infty} a r^n=57 ∑ n = 0 ∞ a r n = 57 and
∑ n = 0 ∞ a 3 r 3 n = 9747 \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 ∑ n = 0 ∞ a 3 r 3 n = 9747 , then
a + 18 r a+18 r a + 18 r is equal to
Step-by-step solution
∑ n = 0 ∞ ar n = 57 a + a r + a r 2 + ∞ = 57 a 1 − r = 57 … ( I ) ∑ n = 0 ∞ a 3 r 3 n = 9747 a 3 + a 3 ⋅ r 3 + a 3 ⋅ r 6 + … … … . … = 9746 a 3 1 − r 3 = 9746 … ( I I ) ( I ) 3 ( I I ) ⇒ a 3 ( 1 − r ) 3 a 3 1 − r 3 = 5 7 3 9717 = 19 a = 19 ∴ a + 18 r = 19 + 18 × 2 3 = 31 \begin{aligned}
& \sum_{n=0}^{\infty} \operatorname{ar}^{\mathrm{n}}=57 \\
& \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 \\
& \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots(I)\\
& \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 \\
& \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots . \ldots=9746 \\
& \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots(II)\\
& \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\frac{\mathrm{a}^3}{(1-\mathrm{r})^3}}{\frac{\mathrm{a}^3}{1-\mathrm{r}^3}}=\frac{57^3}{9717}=19 \\
& \mathrm{a}=19 \\
& \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31 \\
&
\end{aligned} n = 0 ∑ ∞ ar n = 57 a + ar + ar 2 + ∞ = 57 1 − r a = 57 … ( I ) n = 0 ∑ ∞ a 3 r 3 n = 9747 a 3 + a 3 ⋅ r 3 + a 3 ⋅ r 6 + ……… . … = 9746 1 − r 3 a 3 = 9746 … ( II ) ( II ) ( I ) 3 ⇒ 1 − r 3 a 3 ( 1 − r ) 3 a 3 = 9717 5 7 3 = 19 a = 19 ∴ a + 18 r = 19 + 18 × 3 2 = 31 On solving,
r = 2 3 \mathrm{r}=\frac{2}{3} r = 3 2 and
r = 3 2 \mathrm{r}=\frac{3}{2} r = 2 3 (rejected
) ) ) a = 19 ∴ a + 18 r = 19 + 18 × 2 3 = 31 \begin{aligned} & a=19 \\ & \therefore a+18 r=19+18 \times \frac{2}{3}=31\end{aligned} a = 19 ∴ a + 18 r = 19 + 18 × 3 2 = 31 $$
Practice Sequences and Series on Quantrex Academy →
← Previous question If ∑_ r=1 ^n T_r=((2 n-1)(2 n+1)(2 n+3)(2 n+5))/(64) , then _ n arrow ∞ ∑_ r=1…Next question → If ((1)/(α+1)+(1)/(α+2)+… …+(1)/(α+1012))-((1)/(2 · 1)+(1)/(4 · 3)+(1)/(6 · 5)+… .…
More from Sequences and Series
Let α = 3+4+8+9+13+14+… upto 40 terms. If ( β)^ (α)/(1020) is a root of the equation… JEE Main 2026 The sum 1 + (1)/(2)(1^2 + 2^2) + (1)/(3)(1^2 + 2^2 + 3^2) + … upto 10 terms is equal to : JEE Main 2026 The value of 1^3 - 2^3 + 3^3 - … + 15^3 is: JEE Main 2026 The sum of the first ten terms of an A.P. is 160 and the sum of the first two terms of a… JEE Main 2026 For the functions f(θ) = α ^2θ + β ^2θ , and g(θ) = α ^2θ + β ^2θ , α > β > 0 , let _ 0 <… JEE Main 2026 If the sum of the first 10 terms of the series (1)/(1 + 1^4 × 4) + (2)/(1 + 2^4 × 4) +… JEE Main 2026 Let A_1, A_2, A_3, …, A_ 39 be 39 arithmetic means between the numbers 59 and 159 . Then… JEE Main 2026 Let the sum of the first n terms of an A.P. be 3n^2 + 5n . Then the sum of squares of the… JEE Main 2026 ∑_ n=1 ^ 10 ( (528)/(n(n+1)(n+2)) ) is equal to: JEE Main 2026 Let α=(1)/(4)+(1)/(8)+(1)/(16)+…∞ and β=(1)/(3)+(1)/(9)+(1)/(27)+…∞ . Then the value of… JEE Main 2026 The first term of an A.P. of 30 non-negative terms is (10)/(3) . If the sum of this A.P.… JEE Main 2026 Let a_1, a_2, a_3, … be an A.P. and g_1 = a_1, g_2, g_3, … be an increasing G.P. If a_1 =… JEE Main 2026 The sum (1^3)/(1) + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + ⋯ up to 8… JEE Main 2026 Let α, α + 2, α ∈ ℤ , be the roots of the quadratic equation x(x+2) + (x+1)(x+3) +… JEE Main 2026 Let A be the set of first 101 terms of an A.P., whose first term is 1 and the common… JEE Main 2026 If ∑_ k=1 ^ n a_k = 6n^3 , then ∑_ k=1 ^ 6 ( a_ k+1 - a_k 36 )^2 is equal to _______. JEE Main 2026
Full Sequences and Series list
Sequences and Series JEE Maths PYQs JEE Mathematics PYQs
All JEE Main PYQs
