Question

The sum of the first nterims of an A. Pis 63&
the sum of its next 7 terms is 161. Find the 28th
term of this AP
​

Answer 1

ANSWER

Since, S(n) = n/2 [2a + (n-1)]

$given \: s(7) = 63 \\ hence \: s(7) = 7 \div 2(2a + 6d) = 63 \\ or \: 2a + 6d = 18 \: \: ....(1) \\ \\ now \: the \: sum \: of \: 14 \: terms \: is \: \\ s(7) = s(first \: 7) + s(next \: 7) \\ = 63 + 161 \\ = 224 \\ then \: 14 \div 2(2a + 13d) = 32...(2) \\ on \: subtracting \: eq(1) \: and \: (2) \: we \: get \: \\ = (2a + 13d) - (2a + 6d) = 32 - 18 \\ = 7d = 14 \\ d = 14 \div 7 \\ d = 2 \\ puring \: ds \: value \: in \: eq \: (1) \: we \: get \: \\ = 2a + 6d \: = 18 \\ = 2a + 6(2) = 18 \\ 2a + 12 = 18 \\ = 2a = 18 - 12 \\ = 2a = 6 \\ = a = 6 \div 2 \\ = a = 3 \\ since \: a(n) = a + (n - 1)d \\ = a(28) = 3 + 2 \times (27) \\ = a(28) = 57$

Hence, the 28th term of the AP is 57