Let \(\{a_{k}\}\) and \(\{b_{k}\}\), \(k \in \mathbb{N}\), be two G.P.s with common ratio \(r_{1}\) and \(r_{2}\) respectively such that \(\mathrm{a}_{1}=\mathrm{b}_{1}=4\) and \(\mathrm{r}_{1}<\mathrm{r}_{2}\). Let \(\mathrm{c}_{\mathrm{k}}=\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}\), \(\mathrm{k} \in \mathbb{N}\). If \(c_{2}=5\) and \(c_{3}=\frac{13}{4}\) then \(\sum_{\mathrm{k}=1}^{\infty} \mathrm{c}_{\mathrm{k}}-(12 \mathrm{a}_{6}+8 \mathrm{~b}_{4})\) is equal to ______.

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Sequences and Series

Let

{a_{k}}

and

{b_{k}}

k \in N

, be two G.P.s with common ratio

r_{1}

and

r_{2}

respectively such that

a_{1} = b_{1} = 4

and

r_{1} < r_{2}

. Let

c_{k} = a_{k} + b_{k}

k \in N

. If

c_{2} = 5

and

c_{3} = \frac{13}{4}

then

\sum_{k = 1}^{\infty} c_{k} - (12 a_{6} + 8 b_{4})

is equal to ______.

Answer: 9

Since

a_{1} = b_{1} = 4

c_{2} = 4 (r_{1} + r_{2}) = 5

gives

r_{1} + r_{2} = \frac{5}{4}

, and

c_{3} = 4 (r_{1}^{2} + r_{2}^{2}) = \frac{13}{4}

gives

r_{1}^{2} + r_{2}^{2} = \frac{13}{16}

. Then

2 r_{1} r_{2} = (r_{1} + r_{2})^{2} - (r_{1}^{2} + r_{2}^{2}) = \frac{25}{16} - \frac{13}{16} = \frac{3}{4}

, so

r_{1} r_{2} = \frac{3}{8}

; with

r_{1} < r_{2}

this gives

r_{1} = \frac{1}{2}, r_{2} = \frac{3}{4}

. The infinite sums are

\sum a_{k} = \frac{4}{1 - 1/2} = 8

and

\sum b_{k} = \frac{4}{1 - 3/4} = 16

, so

\sum c_{k} = 24

. Also

a_{6} = 4 (\frac{1}{2})^{5} = \frac{1}{8}

and

b_{4} = 4 (\frac{3}{4})^{3} = \frac{27}{16}

, giving

12 a_{6} + 8 b_{4} = \frac{3}{2} + \frac{27}{2} = 15

. Therefore the required value is

24 - 15 = 9

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer