Question

answer this question​

Answer 1

$\huge\boxed{\underline{\bf { \red S \green o \pink L \blue u \orange T \purple i\red O \pink n \green{..}}}}$

Step-by-step explanation:

$so \: using \: formula \: for \: {n}^{th} \: term \: an \: = a_{n} + (n − 1)d$

• a = first term

• n = no. of terms

• d = common difference

So given

⟹a + (3-1) da (7-1)d = 6 2a + 8d = 6

⟹a + 4d = 3

⟹a = 3-4d

⟹(a + 2d)(a + 6d) = 8

⟹(3-2d)(3 + 2d) = 8 {substituting a = 3 - 2d}

⟹9-4d² = 8

⟹d = +1/2 and - 1/2

so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2 a = 3 + 2 = 5

$so \: using \: formula \: for \: S_{n} = \frac{n}{2} [2a(n - 1)d]$

When = + 1/2

$S_{n} = \frac{16}{2} [2 \times 1 + (16 - 1)d]$

$S_{n} = 76$

Similarly when d = - 1/2

Answer 2

$\sf\bold{\underline{Given\::}}$

• $\sf{Sum\:of\:third\:and\:seventh\:term\:of\:an\:AP\:is\:6}$
• $\sf{Product\:of\:third\:and\:seventh\:term\:of\:an\:AP\:is\:8}$

$\sf\bold{\underline{Formula\:used\::}}$

• $\sf{{A}_ {n}=a+(n-1)d}$
• $\sf{{S}_ {n}=\frac{n}{2}(2a+(n-1)d)}$

$\sf\bold{\underline{CASE\: I \::}}$

$\sf\mapsto{{a}_{3} + {a}_{7} = 6}$

$\sf\mapsto{a+2d+a+6d = 6}$

$\sf\mapsto{{2a + 8d= 6}}$

$\sf\mapsto{2(a + 4d) = 6}$

$\sf\mapsto{a + 4d = {\cancel{\frac{6}{2}}}=3}$

$\sf\implies{a = 3 - 4d ------------(i)}$

$\sf\bold{\underline{CASE\: II \::}}$

$\sf\mapsto{{a}_{3} \times {a}_{7} = 8}$

$\sf\mapsto{(a+2d) \times (a+7d) = 8}$

$\sf {Putting\:value \:of\:a\:from\:(i)}$

$\sf\mapsto{(3-4d+2d) \times (3- 4d+7d) = 8}$

$\sf\mapsto{(3-2d) \times (3+2d) = 8}$

$\sf{using\:identity\:=\:(a+b)(a-b)= {a}^2-{b}^{2}}$

$\sf\mapsto{{(3)}^{2} - {(2d)}^{2} = 8}$

$\sf\mapsto{9-4{d}^{2} = 8}$

$\sf\mapsto{4{d}^{2} = 1}$

$\sf\mapsto{{d}^{2} = \frac{1}{4}}$

$\sf\implies{d = +- \frac{1}{2}}$

$\sf\bold{\underline{For\: d\:=\:\frac{1}{2}}}$

$\sf\mapsto{a=3-4d}$

$\sf\mapsto{a=3-4(\frac{1}{2})}$

$\sf\mapsto{a=3-2}$

$\sf\mapsto{a=1}$

$\sf\bold{\underline{For\: d\:=\:\frac{-1}{2}}}$

$\sf\mapsto{a=3-4d}$

$\sf\mapsto{a=3-4(\frac{-1}{2})}$

$\sf\mapsto{a=3+2}$

$\sf\mapsto{a=5}$

Now we will find sum of first sixteen terms of AP,

$\sf\bold{\underline{Condition\:I\::}}$

$\sf\mapsto{When\:a=1\: and\:d=\frac{1}{2}}$

$\sf\mapsto{{S}_ {16}=\frac{16}{2}(2\times1+(16-1)\frac{1}{2})}$

$\sf\mapsto{{S}_ {16}=8(2+\frac{15}{2})}$

$\sf\mapsto{{S}_ {16}=8(\frac{4+15}{2})}$

$\sf\mapsto{{S}_ {16}=8 \times \frac{19}{2}=76}$

$\sf\bold{\underline{Condition\:II\::}}$

$\sf\mapsto{When\:a=5 \:and\:d=\frac{-1}{2}}$

$\sf\mapsto{{S}_ {16}=\frac{16}{2}(2\times5+(16-1)\frac{-1}{2})}$

$\sf\mapsto{{S}_ {16}=8(10-\frac{15}{2})}$

$\sf\mapsto{{S}_ {16}=8(\frac{20-15}{2})}$

$\sf\mapsto{{S}_ {16}=8 \times \frac{5}{2}=20}$

Hence,

when $\sf{a\:=\:1}$ and $\sf{d\:=\:\frac{1}{2}}$, sum of first sixteen terms of AP is 76, and

when $\sf{a\:=\:5}$ and $\sf{d\:=\:\frac{-1}{2}}$, sum of first sixteen terms of AP is 20

@BrainlicaLDoll