Q27 of 74 Page 9

If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.


Given. Sn = 3n2 + 5n

S1 = 3(1)2 + 5 × 1 = 8 ⇒ T1 = 8
S2 = 3(2)2 + 5 × 2 = 22

Thus T1 + T2 = 22

or 8 + T2 = 22

or T2 = 14

Now a = 8, d = 14 – 8 = 6

As Tm = 164
or a +(m – 1)d = 164
or 8 + (m – 1) (6) = 164
or 8 + 6m – 6 = 164
or 6m = 162
or m = = 27
Hence m = 27

More from this chapter

All 74 →
25
Sums of the first p, q, r terms of an A.P are a, b, c respectively. Prove that  

26
The ratio of the sums of m and n terms of an A.P is m2:n2. Show that the ratio of mth and nth term is 2m – 1: 2n – 1.

28
Insert five number between 8 and 26 such that the resulting sequence is an A.P.

29
If (an + bn)/(an-1 + bn-1) is the A.M. between a and b, then find the value of n.