1st term of an a.p is twice its common diff,if the sum of the 1st 10 terms of the a.p is 130 find the 1st term of the a.p

Answer 1

1st term of an a.p is twice its common difference means:

a1 = 2 d

Sum of the first n terms of an A.P.

Sn = n ( a1 + an ) / 2

In this case:

S10 = 10 ∙ ( a1 + a10 ) / 2

130 = 10 ∙ ( a1 + a10 ) / 2

Since:

a10 = a1 + 9 d

130 = 10 ∙ ( a1 + a1 + 9 d ) / 2

130 = ( 10 / 2 ) ∙ ( 2 a1 + 9 d )

130 = 5 ∙ ( 2 ∙ 2 d + 9 d )

130 = 5 ∙ ( 4 d + 9 d )

130 = 5 ∙ 13 d

130 = 65 d

65 d = 130

d = 130 / 65

d = 2

a1 = 2 d

a1 = 2 ∙ 2

a1 = 4