Question

if the ratio of the sum of the first n terms of two A.Ps is (7n+1):(4n+27), then find the ratio of their 9th terms?

Answer 1

Answer:

Hope it helps you!

Step-by-step explanation:

Answer 2

ANSWER:

Given:

• Ratio of the sum of the first n terms of two A.Ps is (7n+1):(4n+27)

To Find:

• Ratio of their 9th terms.

Solution:

$\text{We know that the sum of n terms(S _n ) of an AP is,}\\\\:\longrightarrow S_n=\dfrac{n}{2}(2a+(n-1)d)\\\\\text{We are given that,}\\\\:\implies S_{n_1}:S_{n_2}=(7n+1):(4n+27)\\\\:\implies\dfrac{n}{2}(2a_1+(n-1)d_1):\dfrac{n}{2}(2a_2+(n-1)d_2)=(7n+1):(4n+27)\\\\:\implies\dfrac{\dfrac{n}{2}\!\!\!\!\!\bigg{/}(2a_1+(n-1)d_1)}{\dfrac{n}{2}\!\!\!\!\!\bigg{/}(2a_2+(n-1)d_2)}=\dfrac{7n+1}{4n+27}$

$:\implies\dfrac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\dfrac{7n+1}{4n+27}\\\\\text{On dividing numerator and denominator by 2,}\\\\:\implies\dfrac{\dfrac{2a_1+(n-1)d_1}{2}}{\dfrac{2a_2+(n-1)d_2}{2}}=\dfrac{7n+1}{4n+27}\\\\:\implies\dfrac{\dfrac{2a_1}{2}+\dfrac{(n-1)d_1}{2}}{\dfrac{2a_2}{2}+\dfrac{(n-1)d_2}{2}}=\dfrac{7n+1}{4n+27}\\\\:\implies\dfrac{a_1+\dfrac{(n-1)d_1}{2}}{a_2+\dfrac{(n-1)d_2}{2}}=\dfrac{7n+1}{4n+27}- - - -(1)$

$\text{We know that, general term of an AP(a _n ) is,}\\\\:\implies a_n=a+(n-1)d\\\\\text{So for n = 9,}\\\\:\implies a_9=a+(9-1)d\\\\:\implies a_9=a+(8)d- - - -(2)\\\\\text{So from LHS in (1) and eqn (2),}\\\\:\implies\dfrac{n-1}{2}=8\\\\:\implies n-1=16\\\\:\implies n=17- - - -(3)\\\\\text{Putting value of n from (3) in (1),}\\\\:\implies\dfrac{a_1+\dfrac{(17-1)}{2}d_1}{a_2+\dfrac{(17-1)}{2}d_2}=\dfrac{7(17)+1}{4(17)+27}$

$:\implies\dfrac{a_1+\dfrac{(16)}{2}d_1}{a_2+\dfrac{(16)}{2}d_2}=\dfrac{119+1}{68+27}\\\\:\implies\dfrac{a_1+8d_1}{a_2+8d_2}=\dfrac{120\!\!\!\!\!\!/^{\:\:\:\,24}}{95\!\!\!\!\!/_{\:\:\:\,19}}\\\\\text{From (2)}\\\\:\implies\dfrac{a_{9_1}}{a_{9_2}}=\dfrac{24}{19}\\\\\bf{:\implies(a_{9_1}):(a_{9_2})=24:19}\\\\\text{\bf{Hence, ratio of 9th terms of the APs is 24 : 19.}}$

Formulae Used:

• Sₙ = n/2 [2a + (n - 1)d]
• aₙ = a + (n - 1)d