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The Sum of the First 7 Terms of an A.P. is 63 and the Sum of Its Next 7 Terms is 161. Find the 28th Term of this A.P ? - Mathematics

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Question

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P ?

Solution

Let the first term of the A.P. be a and the common difference be d.
nth term of an A.P. is given by a + (n − 1)d.
∴ 7th term of the A.P. = a + 6d
8th term of the A.P. = a + 7d
14th term of the A.P. = a + 13d

Let the sum of the first seven terms of the A.P. be S1-7.

∴ S1-7 = n2">n2n2(First term + Seventh term)

\[\Rightarrow S_{1 - 7} = \frac{7}{2}\left( a + a + 6d \right)\]
\[ \Rightarrow 63 = 7\left( a + 3d \right) \]
\[ \Rightarrow 9 = a + 3d . . . (1)\]

Let the sum of the next seven terms be S8-14.

∴ S8-14 = n2">n2n2(Eighth term + Fourteenth term)

\[S_{8 - 14} = \frac{7}{2}\left( a + 7d + a + 13d \right)\]
\[ \Rightarrow 161 = 7\left( a + 10d \right)\]
\[ \Rightarrow 23 = a + 10d . . . (2)\]

On subtracting equation (1) from equation (2), we get:
7d = 14
⇒ d = 2
On putting the value of in equation (1), we get:
a = 9 − 6 = 3

\[\text{Now}, {28}^{th} \text{term of the A.P.} = a + 27d = 3 + 27 \times 2 = 57\]
Thus, the 28th term of the given A.P. is 57.
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2013-2014 (March) Foreign Set 3

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