Asked by IFEOMA
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
Answers
Answered by
Bosnian
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
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
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