Question

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

Answer 1

Answer:

The last term is 4.

Step-by-step explanation:

Given :  

A.P is 25,22,19,........,is 116

Here,  a = 25, d = 22  - 25 = - 3, Sn = 116

By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]

116 = n/2 [2 × 25 + (n - 1) × - 3]

116 = n/2 [50 – 3n + 3]

116 = n/2 [53 – 3n]

116 × 2 = n [53 - 3n]

232 = 53n – 3n²

3n² – 53n + 232 = 0

3n² - 24n - 29n + 232 = 0

[By middle term splitting]

3n(n - 8) - 29(n - 8) = 0

(3n - 29) (n - 8) =0

(3n - 29) = 0 or (n - 8) =0

n = 29/3 or n = 8  

n = 29/3 can't be possible because number of terms can't be a fraction.

Therefore, number of terms , n = 8

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

116 = 8/2 [25 + l]

116 = 4 [25 + l]

116/4 = 25 + l

29 = 25 + l

29 - 25 = l

l = 4

Hence, the last term is 4.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Answer:

4

Explanation:

Given

The sum of the terms of A.P.,

25, 22, 19.... is 116

To Find

Last term of the A.P.

Solution

let, the last term is l

The Given A.P. is

25, 22 , 19 .......

Here

a = 25

d = 22-25 = -3

By the identity,

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

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

$116(2) = n [2(25)+(n-1)(-3)]$

$232 = n [50-3n+3]$

$232 = n [53-3n]$

$232 = -3{n}^{2}+53n$

$3{n}^{2}-53n+232 = 0$

Factorise the above quadratic eq.

$3{n}^{2}-24n -29n+232 = 0$

$3n(n-8) -29(n-8) = 0$

$(n-8)(3n-29) = 0$

$n = \dfrac{29}{3}$ is not possible

$\therefore n = 8$

We know that

$\boxed{l = a+(n-1)d}$

$l = 25+(8-1)(-3)$

$l = 25+(7)(-3)$

$l = 25-21$

$\boxed{l = 4}$