Question

please solve this problem​

Answer 1

Answer:

$\rm = 7:16$

Step-by-step explanation:

We know that,

$\boxed{\rm a_n = a +(n-1)d}$

We have,

$\rm \dfrac{a_n}{a'_n} = \dfrac{a+(n-1)d}{a'+(n-1)d'}$

Multiply numerator and denominator by 2,

$\rm = \dfrac{2(a+(n-1)d)}{2(a'+(n-1)d')}$

$\rm = \dfrac{2a+2(n-1)d}{2a'+2(n-1)d'}$

$\rm = \dfrac{2a+(2n-2)d}{2a'+(2n-2)d'}$

$\rm = \dfrac{2a+\{ (2n-1)-1 \} d}{2a'+\{ (2n-1)-1 \} d'}$

$\rm = \dfrac{S_{2n-1}}{S_{2n-1}}$

$\rm \dfrac{S_{2n-1}}{S_{2n-1}} = \dfrac{3n+8}{7n+15}$

Put the value of n as 2n-1

$\rm = \dfrac{3(2n-1)+8}{7(2n-1)+15}$

Put the value of n as 12

$\rm = \dfrac{3(2×12-1)+8}{7(2×12-1)+15}$

$\rm = \dfrac{69+8}{161+15}$

$\rm = \dfrac{77}{176}$

$\rm = \dfrac{7}{16}$