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Question

In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the A.P.


Solution

Let a and b the first term and the common difference of an A.P. respectively.
nth term of an A.P, an=a(n1)d
Sum of n terms of an A.P, Sn=n2[2a+(n1)d]
We have :
Sum of the first 10 terms 102[2a+9d]
210=5[2a+9d]

⇒ 42 = 2a+9d        ………. (1)

15th term from the last = ( 50−15 + 1) th=  36th term from the beginning

Now, a36=a+35d

∴  Sum of the last 15 terms =152 (2a36+(151)d)

                                                 =152 [2(a +35d) +14d]

                                                 =15[a +35d +7d]

⇒2565 = 15[a+42d]

⇒ 171= a+42d                ………….(2)

From (1) and (2), we get,

d = 4

a =  3

So, the A. P. formed is 3, 7, 11, 15  …. and 199.


NCERT Previous Years Papers

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