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If the sum of first n terms of an A.P. is 1 2 (3n2 + 7n), then find its nth term. Hence write its 20th term. - Mathematics

Sum

If the sum of first n terms of an A.P. is  \[\frac{1}{2}\] (3n2 + 7n), then find its nth term. Hence write its 20th term.

 
 
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Solution

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 n terms of an A.P. is \[\frac{1}{2}\] (3n2 + 7n).

∴ First term = =  S= \[\frac{1}{2}\][3(1)2 + 7(1)] = 5.

Sum of first two terms = S=  \[\frac{1}{2}\][3(2)2 + 7(2)] = 13.

 
∴ Second term = S2 − S1 = 13 − 5 = 8.

∴ Common difference = d = Second term − First term
                                           = 8 − 5 = 3
 
Also, nth term = an = a + (n − 1)d
⇒ an = 5 + (n − 1)(3)
⇒ an = 5 + 3n − 3
⇒ an = 3+ 2

Thus, nth term of this A.P. is 3+ 2.

Now, 
a20 = a + (20 − 1)d
⇒ a20 = 5 + 19(3)
⇒ a20 = 5 + 57
⇒ a20 = 62

Thus, 20th term of this A.P is 62.
  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 46 | Page 53
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