Answer :

Let 15th term of the AP series be a15, 7th term be a7, 10th term be a10 and nthterm be an.

According to the question,


15th term of an AP is 3 more than twice its 7th term.


a15 = 3 + 2a7 …(i)


and the a10 = 41 [, 10th term is 41] …(ii)


Now an is given by an = a + (n – 1)d, where a is first term of the AP, n is total number in the series and d is common difference between adjacent numbers in the series.


, a15 = a + (15 – 1)d = a + 14d. Similarly, a7 = a + 6d and a10 = a + 9d


From equation (i),


(a + 14d) = 3 + 2(a + 6d)


a + 14d = 3 + 2a + 12d


14d – 12d = 2a – a + 3


2d = a + 3


2d – a = 3 …(iii)


From equation (ii),


(a + 9d) = 41


9d + a = 41 …(iv)


Adding equations (iii) and (iv), we get


(2d – a) + (9d + a) = 3 + 41


11d = 44


d = 44/11 = 4


d = 4


Putting d = 4 in equation (iii), we get


2d – a = 3


a = 2d – 3


a = 2(4) – 3 = 8 – 3 = 5


Substituting a = 5 and d= 4 in an = a + (n – 1)d,


an = 5 + (n – 1)(4) = 5 + 4n – 4 = 4n + 1


Thus, we have got nth term, that is, 4n + 1.


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