Question

the sum of the 3rd and 7th term of an A.P is 6, the product is 8, find the sum of the first 16 terms of the sequence​

Answer 1

For starter, recall that an arithmetic progression
`\{a_n\}_(n\ge1)` is given recursively by

`\begin{cases}a_1=a\\a_n=a_(n-1)+d&\text{for }n>1\end{cases}`

We can solve this explicitly for
`a_n` in terms of
`a_1`:

`a_n=a_(n-1)+d`

`a_n=(a_(n-2)+d)+d=a_(n-2)+2d`

`a_n=(a_(n-3)+d)+d=a_(n-3)+3d`

and so on, down to

`a_n=a_1+(n-1)d`

More generally, we can express
`a_n` in terms of an arbitrary term
`a_k`:

`a_n=a_k+(n-k)d`

(notice how the index of
`a_k` and the coefficient of
`d` on the right side add to
`n`)

Let
`a_n` be the
`n`-th term of this particular progression. We're given

`\begin{cases}a_3+a_7=6\\a_3a_7=8\end{cases}`

from which it follows that

`a_7=\frac8{a_3}\implies a_3+\frac8{a_3}=6`

`\implies{a_3}^2-6a_3+8=(a_3-4)(a_3-2)=0\implies a_3=4\text{ or }a_3=2`

`\implies a_7=2\text{ or }a_7=4`

If
`a_3=4`, it follows that

`a_7=a_3+(7-3)d\implies2=4+4d\implies d=-\frac12`

Otherwise, if
`a_3=2`, then

`4=2+(7-3)d\implies2=4d\implies d=\frac12`

The sum of the first
`k` terms of an arithmetic progression
`\{a_n\}_(n\ge1)` is

`S_k=\displaystyle\sum_(n=1)^ka_n=\sum_(n=1)^k(a_1+(n-1)d)=ka_1+d\frac{k(k-1)}2`

If
`a_3=4`, then

`a_1=a_3+(1-3)d\implies a_1=5`

so that

`S_(16)=16\cdot5-\frac12\frac{16\cdot15}2=20`

Otherwise, if
`a_3=2`, then

`a_1=a_3+(3-1)d\implies a_1=1`

so that

`S_(16)=16\cdot1+\frac12\frac{16\cdot15}2=76`