Solve the following:
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(a). 27 & (b). 62
Given,
nth term of a sequence an = 4n + 7
On substituting n = 1, 2, 3, 4, and 5, we get the first five terms
a1 = 4 x 1 + 7= 11
a5 = 4 x 5 + 7 = 27
a7 = 4 x 7 + 7 = 35
(a) The 5th term is 27
(b) The sum of 5th and 7th term is
= 5th term + 7 th term
= 27 + 35
= 62
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