Q38 of 40 Page 1

The 17th term of an A.P. is 5 more than twice its 8th term. If 11th terms or A.P. is 43; then find its nth term.

OR


How many terms of A.P. 3, 5, 7, 9, …. Must betaken to get the sum 120?


Given:

a17 = 2a8 + 5


ax = 43


To find: an = ?


ATQ


a17 = 2a8 + 5


a + 16d = 2(a + 7d) + 5


a + 16d = 2a + 19d + 5


–a + 2d = 5 …(1)


a11 = 43


a + 10d = 43 …(2)


from eq. (1) & (2)



Put d = 4 in eq. (1)


–a + 2(4) = 5 ⇒ –a = 5 – 8


a = 3


∴ an = a + (n – 1)d = 3 + (n – 1)4 = 3 + 4


an = 4n – 1


OR


3, 5, 7, 9, ……Sn = 120


a = 3, d = 5 – 3 = 2




240 = n[6 + 2n – 2]


240 = 2n[4 + nn]


120 = n[2 + n]


120 = 2n + n2


n2 + 2n = 120 = 0


n2 + 12x – 10n – 120 = 0


n(n + 12) – 10(n + 12) = 0


(n + 12)(n – 10) = 0


n = –12, n = 10


Clearly, n cannot be negative


∴ neglecting – 12


∴ Number of terms = 10


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