Q. 104.3( 292 Votes )

# Show that a_{1}, a_{2}, ........., a_{n} form an AP where a_{n} is defined as below :

(i) a_{n} = 3 + 4n (ii) a_{n} = 9 - 5n

Also find the sum of the first 15 terms in each case.

Answer :

(i)Let us take different values for a, i.e. 1, 2, 3 and so on

a = 3 + 4 = 7

a_{2} = 3 + 4 x 2 = 11

a_{3} = 3 + 4 x 3 = 15

a_{4} = 3 + 4 x 4 = 19

Here; each subsequent member of the series is increasing by 4 and hence it is an AP.

The sum of "n" terms is given as:

where, n = no. of terms

a= first term

d = common difference

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 7 + (15-1)×4]

= (15/2) [14 + (14)×4]

= (15/2) [70]

= 525

(ii) Let us take values for a, i.e. 1, 2, 3 and so on

a = 9 – 5 = 4

a_{2} = 9 – 5 x 2 = - 1

a_{3} = 9 – 5 x 3 = - 6

a_{4} = 9 – 5 x 4 = - 11

Here; each subsequent member of the series is decreasing by 5 and hence it is an AP.

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 4 + (15-1)×-5]

= (15/2) [8 + (14)×-5]

= (15/2) [8-70]

= 15(-31)

= -465

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The sum of n terms of an A.P. is 3n^{2}+ 5n. Find the A.P. Hence, find its 16th term.