# The sum of first 20 terms of an AP is 400 and sum of first 40 terms is 1600.find the sum of its first 10 terms

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NOW, 20/2 [2a + (20-1)] = 400

10[ 2a + 19d ] = 400

2a + 19d = 40 --------- (i)

similarly, For 40 terms

40/2[ 2a+39d] = 1600

2a + 39d = 80 --------- (ii)

now on subtracting (i) & (ii)

we get, 20d = 40

d = 2

and therefore, 2a+ 19*2 = 40

a = 1

hence sum of 1st ten terms is

10/2[2*1 + (10-1)2]

5[ 2 + 18 ] = 20*5

=

A/q

S20 = (20/2)[2a+(20-1)d]= 400

⇒2a +19d = 40_____(1)

also, S40 =(40/2)[2a+(40-1)d]= 1600

⇒2a +39d = 80____(2)

subtracting (1) and (2), we get

d =2

so, a= 1