Question

The 25th term of an arithmetic sequence is 140 and 27th term is 166.what is the common difference?what is its 35th term?

Answer 1

Given

• 25th term of an AP , $\sf\:a_{25}=140$
• 27th term of an AP , $\sf\:a_{27}=166$

To Find :

• Common difference of an AP
• $\sf\:a_{35}$

Solution :

Let the first term of given AP be a and common difference d .

We know that ,

Genral term of an AP

${\purple{\boxed{\large{\bold{a_n=a+(n-1)d}}}}}$

Given :

• 25th term of an AP is 140

$\sf\:a_{25}=a+(25-1)d$

$\sf\implies\:a_{25}=a+24d$

$\sf\implies\:140=a+24d.....(1)$

And, It's also given that 27th term of an AP is 166.

$\sf\:a_{27}=a+(27-1)d$

$\sf\implies\:a_{27}=a+26d$

$\sf\implies\:166=a+26d.....(2)$

Now subtract equation (1) from (2)

$\sf \: a + 26d = 166 \\ \sf a + 24d = 140 \\ - \: \: \: \: - \: \: = - \\ 0 \: + 2d = 26$

$\sf\:d=\dfrac{26}{2}=13$

Put d = 13 in equation (2)

Then ,

$\sf\:166=a+26\times13$

$\sf\:166=a+338$

$\sf\:166-338=a$

$\sf\:a=-172$

Therefore,

• First term , a= -172
• Common difference = 13

35th term

$\sf\:a_{35}=a+(35-1)d$

$\sf\implies\:a_{35}=-172+34\times13$

$\sf\implies\:a_{35}=-172+442$

$\sf\implies\:a_{35}=270$

Airthemtic progression series :

-172,-159,-146 ......

$\rule{200}2$

More About the topic :

1) Genral term of an Ap

$\sf\:a_n=a+(n-1)d$

2)Sum of n terms of an AP given by :

$\sf \: S_{n} = \dfrac{1}{2}(2a+ (n - 1)d)$