Question

If the sum of a certain number of terms of the A.P. 25, 22, 19, …. is 116. Find the last term.

Answer 1

Given,

An AP whose first term is 25 and the common difference is - 3 

To find : the last term of a given AP 

We know that,

The sum of AP is given by the formula

Solving we get,

3n² - 53n + 232 = 0

n = 8 or n = \(\frac{29}{3}\)

n cannot be a fraction hence we choose n as 8 

Now,

For the finding number of terms, the formula is

Solving we get the last term as 4.

Answer 2

we have,

A.P. is

25,22,19......

Sn​=116

a=25

d=T2​−T1​

d=22−25

d=−3

Then, we know that

Sn​=2n​(2a+(n−1)d)

⇒116=2n​(2×25+(n−1)×(−3))

⇒116=2n​(50−3n+3)

⇒116=2n​(53−3n)

⇒232=n(53−3n)

⇒3n2−53n+232=0

⇒3n2−(29+24)n+232=0

⇒3n2−29n−24n+232=0

⇒n(3n−29)−8(3n−29)=0

⇒(3n−29)(n−8)=0

If

3n−8=0

n=38​(notpossible)

If

n−8=0

n=8

So,

The last term is

Sn​=2n​(a+l)

116=28​(25+l)

116=4(25+l)

116=100+4l

116−100=4l

4l=16

l=4

Hence, the last term of this series is

4.

This is the answer.