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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.

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प्रश्न

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.

योग
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उत्तर

Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\][2a + (n − 1)d

It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161.
∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
                                     = 63 + 161 = 224

Now,
S= \[\frac{7}{2}\][2a + (7 − 1)d]

⇒ 63 = \[\frac{7}{2}\] (2a + 6d)
⇒ 18 = 2a + 6d
⇒ 2a + 6d = 18         ....(1) 

Also,
S14 =  \[\frac{14}{2}\][2a + (14 − 1)d]

⇒ 224 = 7(2a + 13d)
⇒ 32 = 2a + 13d
⇒ 2a + 13d =  32      ....(2) 

On subtracting (1) from (2), we get
13d − 6d = 32 − 18
⇒ 7d = 14
⇒ d = 2
⇒ 2a = 18 − 6d         [From (1)]
⇒ 2a = 18 − 6 × 2
⇒ 2a = 18 − 12
⇒ 2a = 6
⇒ a = 3

Also, nth term = an = a + (n − 1)d
⇒ a28 = 3 + (28 − 1)2
           = 3 + 27 × 2
           = 57

Thus, 28th term of this A.P. is 57.

 

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अध्याय 5: Arithmetic Progressions - Exercise 5.6 [पृष्ठ ५२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progressions
Exercise 5.6 | Q 30 | पृष्ठ ५२

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