the sum of 3rd and 7the term of AP is 6 and their product is 8.Find sum of first 16th term of AP.
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The 3rd term of an AP is a+2d and the 7th term is a+6d. So we have
the sum of 3rd term and 7th term = a+2d+a+6d = 6, or
2a+8d = 6, or
a + 4d = 3 …(1)
Their product is (a+2d)*(a+6d) = 8 …(2)
From (1), a = 3–4d. Put that in (2) to get
(a+2d)*(a+6d) = 8, or
(3–4d+2d)*(3–4d+6d) = 8, or
(3–2d)*(3+2d) = 8, or
9–4d^2 = 8, or
4d^2 = 1, or
d^2 = 1/4, or
d = 1/2.
From (1), a = 3–4d = 3–2 = 1.
The sum of the first 16 terms
S16 = (n/2)[2a+(n-1)d]
= (16/2)[2*1 + (16–1)*1/2]
= 8[2+15/2]
= 8 *19/2
= 76. Answer.
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The 3rd term of an AP is a+2d and the 7th term is a+6d. So we have
the sum of 3rd term and 7th term = a+2d+a+6d = 6, or
2a+8d = 6, or
a + 4d = 3 …(1)
Their product is (a+2d)*(a+6d) = 8 …(2)
From (1), a = 3–4d. Put that in (2) to get
(a+2d)*(a+6d) = 8, or
(3–4d+2d)*(3–4d+6d) = 8, or
(3–2d)*(3+2d) = 8, or
9–4d^2 = 8, or
4d^2 = 1, or
d^2 = 1/4, or
d = 1/2.
From (1), a = 3–4d = 3–2 = 1.
The sum of the first 16 terms
S16 = (n/2)[2a+(n-1)d]
= (16/2)[2*1 + (16–1)*1/2]
= 8[2+15/2]
= 8 *19/2
= 76. Answer.
hope it helps ....
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