Question

Question 19 Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, .

Class X1 - Maths -Sequences and Series Page 193

Answer 1

see attachment ,
actually you have to know that ,
a , b, c , d .... are in GP
A, B, C , D , .... are also in GP .
then,
aA, bB, cC, dD , .... are also in GP.
here we used this concept ( you can see in attachment)

Answer 2

$\huge \underline \bold \color{royalblue} \mathfrak{Answer }$

Series 1 - 2, 4, 8, 16, 32

Series 2 - 128, 32, 8, 2, 1/2

Series formed after product of their corresponding terms - (2×128), (4×32), (8×8), (16×2), (32×1/2)

= 256, 128, 64, 32, 16

The series formed is a GP with a common ratio 1/2

Formula for sum of n terms of a GP

$</em></strong><strong><em>S</em></strong><strong><em>n = \frac{a(1 - {r}^{n}) }{1 - r}$ where r < 1

$</em></strong><strong><em>S</em></strong><strong><em>n = \frac{256(1 - { \</em></strong><strong><em>frac</em></strong><strong><em>{1}{2}</em></strong><strong><em>}</em></strong><strong><em> </em></strong><strong><em>^{5}</em></strong><strong><em>)</em></strong><strong><em> </em></strong><strong><em>}{1 - \frac{1}{2} }$

$</em></strong><strong><em>S</em></strong><strong><em>n = \frac{256 \times 31 \times 2}{32}$

$</em></strong><strong><em>S</em></strong><strong><em>n = 16 \times 31 = \huge \color{green}496$

Hope it helps ✌️ ✌️