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Question
In an A.P. 19th term is 52 and 38th term is 128, find sum of first 56 terms.
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Solution
For an A.P., Let a be the first term and d be the common difference.
t19 = 52 and t38 = 128 ...(Given)
Since tn = a + (n – 1)d
For t19 = 52,
∴ t19 = a + (19 – 1)d
∴ 52 = a + 18d
∴ a + 18d = 52 ...(i)
For t38 = 128,
t38 = a + (38 – 1)d
∴ 128 = a + 37d
∴ a + 37d = 128 ...(ii)
Subtracting equation (i) from (ii), we get,
\[\begin{array}{l}
\phantom{\texttt{0}}\texttt{ a + 37d = 128}\\ \phantom{\texttt{}}\texttt{- a + 18d = 52}\\ \hline\phantom{\texttt{}}\texttt{ (-) (-) (-)}\\ \hline \end{array}\]
∴ 19d = 76
∴ d = 4
Substituting d = 4 in equation (i),
a + 18d = 52
∴ a + 18 × 4 = 52
∴ a + 72 = 52
∴ a = 52 – 72
∴ a = – 20
Now, `S_n = n/2 [ 2a + (n - 1)d]`
∴ S56 = `56/2 [2 × (– 20) + (56 - 1) × 4]`
∴ S56 = 28(– 40 + 220)
∴ S56 = 28 × 180
∴ S56 = 5040
∴ The sum of the first 56 terms is 5040.
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