Question

The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

Answer 1

We know that,
$a_n = a + (n − 1) d$
$a_3 = a + (3 − 1) d$
$a_3 = a + 2d$
Similarly, $a_7 = a + 6d$
Given that,
$a_3 + a_7 = 6$
$(a + 2d) + (a + 6d) = 6$
$2a + 8d = 6$
$a + 4d = 3$
$a = 3 − 4d$ ..…… (i)
Also,
it is given that $(a_3) × (a_7) = 8$
$(a + 2d) × (a + 6d) = 8$
From equation (i),
$(3-4d+2d)(3-4d+6d) = 8$
$(3-2d)(3+2d) = 8$
$9-4d^2 = 8$
$4d^2 = 9-8$
$4d^2 = 1$
$d^2 = \frac 14$
$d = ±\frac 12$
$d = \frac 12$ or $-\frac 12$
From equation number (i),
(Where $d = \frac 12$)
$a = 3-4d$
$a = 3-4(\frac 12)$
$a = 3-2$
$a = -1$
(Where $d =- \frac 12$)
$a = 3-4(-\frac 12)$
$a = 3+2$
$a = 5$
$S_n = \frac n2[2a+(n-1)d]$
(Where $a = 1$ and $d =\frac 12$)
$S_{16} = \frac {16}{2}[2(1)+(16-1)(\frac 12)]$

$S_{16} = 8[2+\frac {15}{2}]$
$S_{16} = 4 \times 19$
$S_{16} = 76$
(where $a = 5$and $d =- \frac 12$)
$S_{16} = \frac {16}{2}[2(5)+(16-1)(-\frac 12)]$

$S_{16} = 8[10+15(-\frac 12)]$
$S_{16}= 8 \times \frac 52$
$S_{16} = 20$