32) The 13th term of an A.P is 3 and the sum of the first 13 terms is 234. Find the common difference and the sum of first 21 terms.
Answer:
Given the 13th term = 3, so \(t_{13} = a + 12d = 3\) ..... (1)
Sum of first 13 terms = 234, so \(S_{13} = \frac{13}{2}[2a + 12d] = 234\)
\(13(a + 6d) = 234 \implies a + 6d = 18\) ..... (2)
Solving (1) and (2):
(1) - (2) \(\implies (a+12d) - (a+6d) = 3 - 18\)
\(6d = -15 \implies d = -\frac{15}{6} = -\frac{5}{2}\)
Substitute d in (2): \(a + 6(-\frac{5}{2}) = 18 \implies a - 15 = 18 \implies a = 33\)
Common difference \(d = -\frac{5}{2}\)
Sum of first 21 terms, \(S_{21} = \frac{21}{2}[2a + (21-1)d]\)
\(S_{21} = \frac{21}{2}[2(33) + 20(-\frac{5}{2})] = \frac{21}{2}[66 - 50] = \frac{21}{2}[16] = 21 \times 8 = 168\).