Question

The last term of a series in A.P. is 40 , The sum is 952 and the common difference is -2 . Find the first term and the number of term in the series.




Chapter name is Arithmetic Progression

Answer 1

Hi,

Let ' a ' be the first term and ' d ' be

the common difference and last term

be ' l ' of an A.P.,

a + ( n - 1 ) d = l ------( 1 )

Last term = l = 40 given

a + ( n - 1 ) ( -2 ) = 40

a -2n + 2 = 40

a - 2n = 38
a = 38 + 2n ------( 2 )

We know that ,
__________________________

Sum of first ' n ' terms be Sn.

Sn = n (a + l ) /2 ------( 3 )
__________________________
According to the problem given,

Sn = 952,

l = 40,

d = -2

Put above values in ( 3 ), we get

952 = n ( a + 40 ) /2

952 = n ( 38 + 2n + 40 )/2 [ from ( 2 ) ]

952 = 2n ( 19 + n + 20 ) /2


952 = n ( 39 + n )

0 = 39n + n^2 - 952

n^2 + 39n -952 = 0

n^2 + 56n -17n - 952 = 0

n ( n + 56 ) - 17 ( n + 56 ) = 0

( n + 56 ) ( n - 17 ) = 0

Therefore ,

n + 56 = 0 or n - 17 = 0

n = - 56 or n = 17

Put n values in equation ( 2 ) ,

We get ' a ' values

a = 38 + 2n

If n = -56 then a = -74

If n = 17 then a = 72

Therefore,

1 ) If first term a = -74 and

number of terms should be n= -56,

2 ) If first term a = 72 and

number of term should be n = 17

In both conditions sum of the terms = 952


I hope this helps you.

:)