Question

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

sixteen terms of the AP.
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Answer 1

I hope it will help you.

Answer 2

$\huge\bf Question:-$

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

$\huge \bf \: Solution:-$

Let the first term and the common difference of

the given AP be a and d, respectively.

According to the question,

Third term + Seventh term = 6

$\implies \: \: \: \bf \: [a \: + (3 - 1) \: d \: + \: a + (7 - 1) \: d ]= 6$

$\: \: \: \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [\because \: a_n \: = a \: + \:( n - 1) \: d]$

$\implies \: \: \: \: \: \: \: \: \: \: \: \: \bf \: \: \: (a \: + \: 2d) \: + \: (a \: + \: 6d) = 6$

$\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: 2a + 8d = 6$

$\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: a \: + \: 4d = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)$

$\bf \: and \: third \: term \times Seventh \: term \: = 8$

$\implies \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \:( a \: + \: 2d)(a \: + \: 6d) =8$

$\implies \: \: \: \bf [(a + 4d) - 2d] \: [(a + 4d) + 2d] = 8$

$\implies \: \: \: \: \: \bf \: (3 - 2d)(3 + 2d) = 8 \: \: \: \: \: \: \: [using \: Eq. \: (i)]$

$\implies \: \: \: \: \: \: \: \bf \: 9 - 4d {}^{2} = 8 \: \: \: \: \: \: \: \: [ \because \: (a - b) \: (a + b) = a {}^{2} - b {}^{2} ]$

$\implies \: \: \: \: \: \: \: \: \: \: \bf \: 4d {}^{2} = 9 - 8$

$\implies \: \: \: \: \: \: \: \: \: \: \: \bf d {}^{2} = \: \frac{1}{4}$

$\implies \: \: \: \: \: \: \: \: \: \: \bf d = \: ± \frac{1}{2}$

$\bf \: When, \: d \: = \: \frac{1}{2} ,\: then \: from \: Eq. \: (i), \: we \: get$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: a + 4 \bigg( \frac{1}{2} \bigg) = 3$

$\implies \: \: \: \bf a + 2 = 3$

$\implies \: \: \: \bf a = 3 - 2$

$\implies \: \: \: \bf a = 1$

$\bf \: Now, \: sum \: of \: first \: sixteen \: terms \: of \: the \: AP,$

$\bf \: s_16 \: = \: \frac{16}{2} \: [2a + (16 - 1) \: d] \: \: \: \: \bigg [ \because \: s_n \: = \: \frac{n}{2} [2a + (n - 1) \: d] \bigg]$

$\bf = 8 \: [2a + 15d] = 8 \bigg [2(1) + 15 \bigg( \frac{1}{2} \bigg) \bigg] \: \: \: \: \bigg [ \because \: a = 1 \: d = \frac{1}{2} \bigg]$

$\bf \: = \: 8 \bigg(2 + \frac{15}{2} \bigg) = \: 8 \bigg( \frac{19}{2} \bigg) = 76$

$\bf \: When, \: d \: = - \frac{1}{2} \: then \: from \: Eq. \: (i) ,\: we \: get$

$\bf \: a + 4 \bigg(- \frac{1}{2} \bigg) = 3 \implies \: a - = 3 \implies \: a = 5$

$\bf \: Now, \: sum \: of \: first \: sixteen \: terms \: of \: this \: AP, \:$

$\bf \: s_16 \: = \: \frac{16}{2} \: [2a + (16 - 1) \: d] \: \: \: \: \: \: \: \bigg [ \because \: \: s_n \: = \: \frac{n} {2} \: [2a + (n - 1) \: d] \bigg ]$

$\bf \: = 8 \: (2a + 15d)$

$\bf \: = 8 \bigg [2(5) + 15 \bigg( - \frac{1}{2} \bigg) \bigg] = 8 \bigg(10 - \frac{15}{2} \bigg) = 8 \bigg( \frac{5}{2} \bigg) = 20$

$\bf \: Hence, \: sum \: of \: first \: sixteen \: terms, \: s_1_6 \: \bf{20} \: or \: \bf{76}.$