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

Asked by Pragya Singh | 1 year ago |  106

##### Solution :-

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

Sn = $$\dfrac{n}{2}$$ [2a + (n-1)d] = 116

116 = $$\dfrac{n}{2}$$ [2(25) + (n-1)(-3)]

116 x 2 = n [50 – 3n + 3]

232 = n [53 – 3n]

232 = 53n – 3n2

3n2 – 53n + 232 = 0

3n2 – 24n – 29n+ 232 = 0

3n(n – 8) – 29(n – 8) = 0

(3n – 29) (n – 8) = 0

Hence,

n =$$\dfrac{29}{3}$$ or n = 8

As n can only be an integral value, n = 8

Thus, 8th term is the last term of the A.P.

a8 = 25 + (8 – 1)(-3)

= 25 – 21

= 4

Answered by Abhisek | 1 year ago

### Related Questions

#### Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.

#### Find the two numbers whose A.M. is 25 and GM is 20.

Find the two numbers whose A.M. is 25 and GM is 20.

#### If a is the G.M. of 2 and 1/4 find a.

If a is the G.M. of 2 and $$\dfrac{1}{4}$$ find a.

#### Find the geometric means of the following pairs of numbers

Find the geometric means of the following pairs of numbers:

(i) 2 and 8

(ii) a3b and ab3

(iii) –8 and –2

Insert 5 geometric means between $$\dfrac{32}{9}$$ and $$\dfrac{81}{2}$$.