Question

if the sum of the terms of the series 25,22,19,..............is 116 then find the last term and the number of terms

Answer 1

it's an AP with common difference of -3 so we get s=n/2(2a + (n-1)×d) by this formula the n which is number of terms and by putting n in a+(n-1)d we get the last term

Answer 2

Hi ,

Given series is 25 , 22 , 19 , ... is an AP 

first term = a = 25 

common difference = d = a2 - a1 

d = 22 - 25 = -3 

last term = a + ( n - 1 ) d 

l = 25 + ( n - 1 ) ( -3 )

 = 25 - 3n + 3

= 28 -3n

sum of n terms = 116 

Sn = n /2 [ a + l ]

n / 2 [ 25 + 28 - 3n ] = 116 

n ( 53 - 3n ) = 232

53n - 3n² = 232

3n² - 53n + 232 =0 

3n² - 24n - 29n + 232 =0 

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

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

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

n = 8 or 3n = 29 

n = 8 or n = 29 / 3

but n should be a natural number .

∴ n = 8 

last term = l = 28 - 3n

l = 28 - 3 × 8

l = 28 - 24 

l = 4 

number of terms = n = 8 

last term = l = 4

I hope this helps you .

: )