If 100 times the 100th term of an AP with non-zero common difference equal to the 50 times its 50th term, then the 150th term of AP is

Solution :

Let a be the first term and d (d \(\ne\) 0) be the common difference of the given AP, then

\(T_{100}\) = a + (100 – 1)d = a + 99d

\(T_{50}\) = a + (50 – 1)d = a + 49d

\(T_{150}\) = a + (150 – 1)d = a + 149d

Since 100 times the 100th term = 50 times its 50th term

\(\implies\)  100(a + 99d) = 50(a + 49d)

\(\implies\)  2a + 198d = a + 49d

\(\implies\)  a + 149d = 0

\(\therefore\)   \(T_{150}\) = 0


Similar Questions

Let \(a_n\) be the nth term of an AP. If \(\sum_{r=1}^{100}\) \(a_{2r}\) = \(\alpha\) and \(\sum_{r=1}^{100}\) \(a_{2r-1}\) = \(\beta\), then the common difference of the AP is

A man saves Rs 200 in each of the first three months of his service. In each of the subsequent months, his saving increases by Rs 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs 11040 after

Find the sum of n terms of the series 1.3.5 + 3.5.7 + 5.7.9 + ……

If x, y and z are in AP and \(tan^{-1}x\), \(tan^{-1}y\) and \(tan^{-1}z\) are also in AP, then

The sum of first 20 terms of the sequence 0.7, 0.77, 0.777, ……. , is

Leave a Comment

Your email address will not be published.