Question

5 29. The sum of the first 20 terms in progression is 200 and the sum of the next 10 terms is arithmetic 400. Find the fortieth term.A) 49 C) 69 B) 59 D) 79 ​

Answer 1

Given,

The sum of the first $20$ terms in progression is $200$ and the sum of the next $10$ terms is arithmetic $400$.

Formula,

$S_{n}=\frac{n}{2}[2a+(n-1)d]$

Step1:

$n=20$, $S_{n} =200$

$200=\frac{20}{10}[2a+(20-1)d]\\10[2a+19d]=200\\2a+19d=20--------(1)$

Step2:

Sum of the next $10$ terms

So, total terms are $30$

$\frac{30}{2}[2a+29d]-\frac{20}{2}[2a+19d]=400\\15[2a+29d]-10[2a+19d]=400\\5[(3(2a+29d))-2(2a+19d)]=400\\6a+87d-4a-38d=80\\2a+49d=80--------(2)$

Subtract equation (2)to equation(1)

$2a+49d-2a-19d=80-20\\30d=60\\d=2$

Substitute $d=2$ in equation (2)

$2a+49\times2=80\\2a+98=80\\2a=80-98\\2a=18\\a=-9$

$T_{40}=a+(40-1)d$

$T_{40}=-9+39\times2\\T_{40}=-9+78\\T_{40}=69$

Therefore, the fortieth term is $69$.