Let a1, a2, a3, . . . . . . . , an, . . . . . be in A.P. If a3 + a7 + a11 + a15 = 72, then the sum of its first 17 terms is equal to :

Sequences and Series question

2016 · 10 Apr · Shift 1 · Q21

2016 · 10 Apr · Shift 1 · Q21

View written solutionFree

Correct answer: A

Let the A.P. have first term $a$ and common difference $d$ .

Then $a_n=a+(n-1)d$
Write the given terms:
$\quad a_7=a+6d, \quad a_{11}=a+10d, \quad a_{15}=a+14d$$ Their sum is given as $72$: $$ (a+2d)+(a+6d)+(a+10d)+(a+14d)=72 $$ $$4a+32d=72$$ $$a+8d=18$$$
Notice that $a+8d=a_9$ . So, $a_9=18$
We need the sum of first $17$ terms.

For an odd number of terms in an A.P., the sum equals: $S_{17}=17\times \text{middle term}=17a_9$

Hence, $S_{17}=17\times 18=306$
Therefore, the correct option is: $\boxed{306}$
Comparing with the stored correct answer:

Stored answer = A = $306$

This matches our result.

2016 · 10 Apr · Shift 1 · Q21

2016 · 10 Apr · Shift 1 · Q21

View written solutionFree

Correct answer: A

Let the A.P. have first term $a$ and common difference $d$ .

Then $a_n=a+(n-1)d$
Write the given terms:
$\quad a_7=a+6d, \quad a_{11}=a+10d, \quad a_{15}=a+14d$$ Their sum is given as $72$: $$ (a+2d)+(a+6d)+(a+10d)+(a+14d)=72 $$ $$4a+32d=72$$ $$a+8d=18$$$
Notice that $a+8d=a_9$ . So, $a_9=18$
We need the sum of first $17$ terms.

For an odd number of terms in an A.P., the sum equals: $S_{17}=17\times \text{middle term}=17a_9$

Hence, $S_{17}=17\times 18=306$
Therefore, the correct option is: $\boxed{306}$
Comparing with the stored correct answer:

Stored answer = A = $306$

This matches our result.