The sum of the third and seventh terms of an ap is 40 and the sum of its sixth and fourteenth terms is 70. Find the sum of the first ten terms of an ap
Answers
a₃ = a₁ + 2d
a₇ = a₁ + 6d
a₃ + a₇ = 2a₁ + 8d = 40
a₁ + 4d = 20..............................(1)
a₆ = a₁ + 5d
a₁₄ = a₁ + 13d
a₆ + a₁₄ = 2a₁ + 18d = 70
a₁ + 9d = 35..............................(2)
solving 1, 2
a₁ = 8, d= 3
s₁₀ = a₁+a₂+a₃+a₄+...+a₁₀
= 5(2a₁ + 9d) = 5 (16+27) = 215
Given : sum of the third and the seventh terms of an ap is 40 and the sum of its sixth and 14th term is 70
To Find : the sum of the first 10 terms.
Solution:
Let say AP
a , a + d , a + 2d + .............
first term = a
common difference = d
an = a + (n-1)d
3rd term = a + 2d
7th term = a + 6d
sum of the third and the seventh terms of ap is 40
=> a + 2d + a + 6d = 40
=> a + 4d = 20 ..Eq1
6th term = a + 5d
14th term = a + 13d
a + 5d + a + 13d = 70
=> a + 9d = 35 Eq2
Eq2 - Eq1
=> 5d = 15
=> d = 3
a + 4d = 20 => a + 12 = 20 => a = 8
a = 8 d = 3
Sn = (n/2)(2a + (n-1)d)
=> S10 = (10/2)(2 * 8 + (10 - 1) 3)
= 5 ( 15 + 27)
= 5 * 43
= 215
sum of the first 10 terms. = 215
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