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 A.P.​

Answer 1

AnswEr :

$\bigstar\:\large\boxed{\sf T_n = a + (n-1)d}\\\qquad\qquad\qquad \qquad \sf where :\\\bullet\:\textsf{a = First Term}\\\bullet\:\textsf{n = Number of Terms}\\\bullet\:\textsf{d = Common Difference}$

$\rule{120}{1}$

$\underline{\bigstar\:\textsf{Sum of 3rd and 7th term of AP :}}$

$:\implies\tt T_3 + T_7 = 6\\\\\\:\implies\tt a + (3-1)d + a + (7-1)d = 6\\\\\\:\implies\tt a + 2d + a + 6d = 6\\\\\\:\implies\tt 2a + 8d = 6\\\\\\:\implies\tt 2(a + 4d) = 6\\\\\\:\implies\tt a + 4d = 3\\\\\\:\implies\tt a = 3 - 4d \qquad \dfrac{\quad}{} \:eq.(1)$

$\underline{\bigstar\:\textsf{Product of 3rd and 7th term of AP :}}$

$:\implies\tt T_3\times T_7 = 8\\\\\\:\implies\tt (a+2d)(a+6d)=8\\\\{\scriptsize\qquad\bf{\dag}\:\:\textsf{Putting the value of a from eq.(1) :}}\\\\:\implies\tt (3-4d+2d)(3-4d+6d)=8\\\\\\:\implies\tt (3 - 2d)(3 + 2d) = 8\\\\{\scriptsize\qquad\bf{\dag}\:\:\sf{(a-b)(a+b)=a^2-b^2}}\\\\:\implies\tt (3)^{2} -(2d)^{2} = 8\\\\\\:\implies\tt 9 - 4d^{2} = 8\\\\\\:\implies\tt 9 - 8 = 4d^{2}\\\\\\:\implies\tt 1 = 4{d}^{2}\\\\\\:\implies\tt\dfrac{1}{4} = d^{2}\\\\\\:\implies\tt \sqrt{\dfrac{1}{4}} = d\\\\\\:\implies\tt d =\pm\:\dfrac{1}{2}$

$\rule{200}{2}$

$\underline{\bigstar\:\textsf{When d is equal to \sf\frac{1}{2} :}}$

$\dashrightarrow\:\:\tt S_{16} = \dfrac{n}{2} \times \lbrack 2a + (n-1)d\rbrack \\\\\\\dashrightarrow\:\:\tt S_{16} = \dfrac{16}{2} \times \lbrack 2(3 - 4d) + (16-1)d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \lbrack 6 - 8d + 15d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \lbrack 6 + 7d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \bigg\lbrack 6 +\dfrac{7}{2} \bigg\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times\dfrac{(12 + 7)}{2}\\\\\\\dashrightarrow\:\:\tt S_{16} =4 \times 19\\\\\\\dashrightarrow\:\:\underline{\boxed{\tt S_{16} =76}}$

$\underline{\bigstar\:\textsf{When d is equal to \sf - \frac{1}{2} :}}$

$\dashrightarrow\:\:\tt S_{16} = \dfrac{n}{2} \times \lbrack 2a + (n-1)d\rbrack \\\\\\\dashrightarrow\:\:\tt S_{16} = \dfrac{16}{2} \times \lbrack 2(3 - 4d) + (16-1)d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \lbrack 6 - 8d + 15d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \lbrack 6 + 7d\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times \bigg\lbrack 6 - \dfrac{7}{2} \bigg\rbrack\\\\\\\dashrightarrow\:\:\tt S_{16} = 8\times\dfrac{(12 - 7)}{2}\\\\\\\dashrightarrow\:\:\tt S_{16} =4 \times 5\\\\\\\dashrightarrow\:\:\underline{\boxed{\tt S_{16} =20}}$

⠀

$\therefore\:\underline{\textsf{Sum of sixteen terms can be \textbf{76 or 20.}}}$

Answer 2

$\bf{\large{\boxed{\sf{ANSWER\::}}}}}$

$\bf{\large{\underline{\underline{\bf{Given\::}}}}}$

The sum of the third and seventh terms of an A.P. is 6 and their products is 8.

$\bf{\large{\underline{\underline{\bf{To\:find\::}}}}}$

The sum of first sixteen terms of the A.P.

$\bf{\Large{\underline{\underline{\tt{\red{Explanation\::}}}}}}$

We know that Formula of an A.P;

$\bf{\large{\boxed{\sf{a_{n}=a+(n-1)d}}}}$

$\bf{\large{\underline{\underline{\tt{\red{A.T.Q\::}}}}}}$

$\longrightarrow\tt{a_{3}+a_{7}=6}\\\\\\\longrightarrow\tt{a+2d+a+6d=6}\\\\\\\longrightarrow\tt{2a+8d=6}\\\\\\\longrightarrow\tt{2(a+4d)=6}\\\\\\\longrightarrow\tt{a+4d=\cancel{\dfrac{6}{2}} }\\\\\\\longrightarrow\tt{a+4d=3}\\\\\\\longrightarrow\tt{\pink{a=3-4d.............................(1)}}$

$\blacksquare$ Their Products :

$\longrightarrow\tt{a_{3}\times a_{7}=8}\\\\\\\longrightarrow\tt{(a+2d)\times (a+6d)=8}\\\\\\\longrightarrow\tt{a^{2} +6ad+2ad+12d^{2} =8}\\\\\\\longrightarrow\tt{a^{2} +8ad+12d^{2} =8}\\\\\\\longrightarrow\tt{(3-4d)^{2} +8(3-4d)d+12d^{2} =8\:\:\:\:\:\:\big[From(1)\big]}\\\\\\\longrightarrow\tt{9-16d^{2} -24d+24-32d^{2} +12d^{2} =8}\\\\\\\longrightarrow\tt{d^{2} (16-32+12)+d\cancel{(-24+24)}=-1}\\\\\\\longrightarrow\tt{d^{2} (28-32)=-1}\\\\\\\longrightarrow\tt{-4d^{2} =-1}$

$\longrightarrow\tt{\cancel{-}4d^{2} =\cancel{-}1}\\\\\\\longrightarrow\tt{d^{2} =\dfrac{1}{4} }\\\\\\\longrightarrow\tt{d\:=\:\sqrt{\frac{1}{4} } }\\\\\\\longrightarrow\tt{\pink{d\:=\:\pm\dfrac{1}{2} }}$

$\blacksquare$ Putting the value of d (+ve) in equation (1), we get;

$\mapsto\tt{a=3-4d}\\\\\\\mapsto\tt{a=3-\cancel{4}*\dfrac{1}{\cancel{2}} }\\\\\\\mapsto\tt{a=3-2}\\\\\\\mapsto\tt{\orange{a\:=\:1}}$

$\blacksquare$ Putting the value of d (-ve) in equation (1), we get;

$\mapsto\tt{a\:=\:3-4d}\\\\\\\mapsto\tt{a=3-4*-\dfrac{1}{2}} \\\\\\\mapsto\tt{a=3-(-\cancel{\dfrac{4}{2} )}}\\\\\\\mapsto\tt{a=3-(-2)}\\\\\\\mapsto\tt{a=3+2}\\\\\\\mapsto\tt{\orange{a=5}}$

$\blacksquare$ The sum of sixteen terms of the A.P. (+ve) d :

$\dashrightarrow\tt{S_{n}=\dfrac{n}{2} \big[2a+(n-1)d\big]}\\\\\\\dashrightarrow\tt{S_{16}=\cancel{\dfrac{16}{2}} \bigg[2(1)+(16-1)*\dfrac{1}{2} \bigg]}\\\\\\\dashrightarrow\tt{S_{16}=8\bigg[2+15*\dfrac{1}{2} \bigg]}\\\\\\\dashrightarrow\tt{S_{16}=8*\dfrac{4+15}{2} }\\\\\\\dashrightarrow\tt{S_{16}=\cancel{8}*\dfrac{19}{\cancel{2}} }\\\\\\\dashrightarrow\tt{S_{16}=4*19}\\\\\\\dashrightarrow\tt{\blue{S_{16}=76}}$

$\blacksquare$ The sum of sixteen terms of the A.P. (-ve) d :

$\dashrightarrow\tt{S_{n}=\dfrac{n}{2} \big[2a+(n-1)d\big]}\\\\\\\dashrightarrow\tt{S_{16}=\cancel{\dfrac{16}{2}} \bigg[2(5)+(16-1)*-\dfrac{1}{2} \bigg]}\\\\\\\dashrightarrow\tt{S_{16}=8\bigg[10+15*-\dfrac{1}{2} \bigg]}\\\\\\\dashrightarrow\tt{S_{16}=8*\dfrac{20-15}{2} }\\\\\\\dashrightarrow\tt{S_{16}=\cancel{8}*\dfrac{5}{\cancel{2}} }\\\\\\\dashrightarrow\tt{S_{16}=4*5}\\\\\\\dashrightarrow\tt{\blue{S_{16}=20}}$

∴ The sum first sixteen terms of the A.P. is 20 or 76.