Question

If the third term of an ap is 5 and the ratio of it's 6th term to the tenth term is 7:13, find the sum of first 20 terms

Answer 1

Answer:

Given :

T3 = 5

T6 : T10 = 7:13

T3 = a + ( 3-1) d = 5

= a + 2d = 5

= 2d = 5 - a

= d = 5 - a / 2 (eq1)

T6 : T10 = 7 : 13

T6 / T10 = 7 / 13

a + ( 6-1 ) d / a + ( 10-1) = 7 / 13

a + 5d / a + 9d = 7 / 13

13 ( a + 5d ) = 7 ( a + 9d)

13a + 65d = 7a + 63d

13a - 7a = 63d - 65d

6a = -2d

6a = -2 ( 5 - a /2) ( from eq 1)

6a = -5 + a

6a -a = -5

5a = -5

a = -1

Now, putting the value of a in eq 1

d = 5 -(-1) /2

= 5 + 1 / 2

= 6/2

= 3

a = -1 , d = 3

Now the sum of first 20 terms

S20 = n/2 [ 2a + ( n -1 ) d]

=20/2 [ 2 (-1) + (20 -1 ) 3

= 10 [ -2 + 19×3]

= 10 [ -2 + 57]

= 10 [ 55]

= 10 × 55

= 550

So, S20 = 550