Question

If twelfth term of an A.P. is 30, then the sum of its third, twelfth and twenty-first term is

a)30


b) 90


c) 60


d) 45

Answer 1

Answer:

A12 = 30

=> A + 11d = 30 => A = 30 - 11d

A3 + A12 + A21 = A + 2d + A + 11d + A + 20d

= 3A + 33d

= 3 ( A + 11d )

= 3 ( A12 ). ( A + 11d = A12 )

= 3 ( 30 )

= 90

Answer 2

The sum of the third, twelfth, and twenty-first terms is b)90.

Step-by-step explanation:

Given:

$t_{12}$= 30, It is an A.P.

To find:

sum of $t_{3} \ ,t_{12} \ ,t_{21}$

Solution:

The given terms are in A.P.

As we know, a term (t) in an A.P.= a+d

In this example $t_{12}$= a+ $t_{11}$ (d)

$t_{12}$= a+11d

Now, similarly  $t_{3}$= a+2d

$t_{21}$= a+20d

Sum of these terms= (a+2d)+(a+11d)+(a+20d)

$t_{3}$+ $t_{12}$+ $t_{21}$= a+2d+a+11d+a+20d

$t_{3}$+ $t_{12}$+ $t_{21}$= 3a+33d..........after adding the like terms

$t_{3}$+ $t_{12}$+ $t_{21}$= 3(a+11d).....(taking 3 as the common factor)

But, $t_{12}$= a+11d= 30.....(given)

∴  $t_{3}$+ $t_{12}$+ $t_{21}$= 3(30)

$t_{3}$+ $t_{12}$+ $t_{21}$=90

Thus the sum of the third, twelfth, and twenty-first terms will be 90.

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