Question

find the sum of the first 25 terms of the A.P. whose 5th term is 34 and 9th term is 58.​

Answer 1

Given :-

5th term is 34

9th term is 58

Solution :-

a5 = a + ( n - 1)d = 34

a + ( 5 - 1 )d = 34

a + 4d = 34 .............eq( 1)

a9 = a + ( 9 - 1 ) d = 58

a + 8d = 58 ..............eq( 2)

Now , Subtract eq( 2) from eq( 1 )

a + 8d - ( a + 4d ) = 58 - 34

a + 8d -a - 4d = 24

4d = 24

d = 24/4

d = 6

Hence , The common difference is 6

Now , put the value of d in eq( 1 )

a + 4d = 34

a + 4 * 6 = 34

a + 24 = 34

a = 34 - 24

a = 10

The first term of an AP is 10

Now , The sum of first 25 terms

Sn = n/2 ( 2a + ( n - 1)d )

S25 = 25/2 ( 2 * 10 + ( 25 - 1) *6)

S25 = 25/2 ( 20 + 144)

S25 = 25/2 ( 164)

S25 = 25 * 82

S25 = 2050

Hence , The sum of 25th terms of an AP is 2050

Answer 2

Answer:

1050 is the correct answer