Which term of the AP 3, 15, 27, 39,… will be 120 more than its 21st term?

Explanation

Here a= 3 and = (15-3) = 12.
`therefore ` 21 st term is given by
`T_(21) = a +(21-1)d = a+ 20d = (3 + 20 xx 12) = 243.`
Required term = (243 + 120) = 363.
Let it be nth term. Then,
`T_(n) = 363 rArr a + (n-1)d = 363`
`rArr 3 + (n-1) xx 12 = 363`
`rArr 12n = 372 rArr n = 31`
Hence, 31st term is the required term.