Question

The sum of the first 3 terms in an AP is 51 and that of the last 3 is 99. If the AP has 11 terms, help Joe find the first number of the series.

• A. 17
• B. 39
• C. 15
• D. 2

Answer 1

The correct option is C 15

Let the first three terms be a, (a + r), and (a + 2r), where r is the common difference.

The last three terms are (a + 8r), (a + 9r), and (a + 10r).

By the given condition,
$a+a+r+a+2r=51$
or, $3a+3r=51$
or, $a+r=17$

Again,
$(a+8r)+(a+9r)+(a+10r)=99$

$⟹\$3a+27r=99$

$⟹51+24r=99$

$⟹24r=99–51$

$⟹24r=48$

$⟹r=\frac{48}{24}=2$

$a+r=17$
$⟹a=15$