Question

the last term of a series in AP is 40 the sum is 952 and the common difference is -2 find the first term and the last term in the series

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

First term of an AP is a and number of terms be n.

Now, given that,

The last term of a series in AP is 40 and the common difference is -2.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm :\longmapsto\:40 = a + (n - 1)( - 2)$

$\rm :\longmapsto\:40 = a - 2n + 2$

$\bf\implies \:a = 38 + 2n - - - - (1)$

Also, given that,

The last term of a series in AP is 40, the sum is 952 and the common difference is -2.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

So, on substituting the values, we get

$\rm :\longmapsto\:S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)$

$\rm :\longmapsto\:952\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:( - 2) \bigg)$

$\rm :\longmapsto\:952\:=\dfrac{n}{2} \bigg(2 \:a\: - \:2n\: + \:2\: \bigg)$

$\rm :\longmapsto\:952\:=n \bigg(\:a\: - \:n\: + \:1\: \bigg)$

$\rm :\longmapsto\:952\:=n \bigg(\:38 + 2n\: - \:n\: + \:1\: \bigg)$

$\rm :\longmapsto\:952\:=n \bigg(\:39 + n\: \bigg)$

$\rm :\longmapsto\:952\:= {n}^{2} + 39n$

$\rm :\longmapsto\:{n}^{2} + 39n - 952 = 0$

$\rm :\longmapsto\:{n}^{2} + 56n - 17n - 952 = 0$

$\rm :\longmapsto\:n(n + 56) - 17(n + 56) = 0$

$\rm :\longmapsto\:(n + 56)(n - 17) = 0$

$\bf\implies \:n = 17 \: \: \: or \: n = - \: 56 \: \{rejected \}$

Substituting the value of n in equation (1), we get

$\rm :\longmapsto\:a = 38 + 2 \times 17$

$\rm :\longmapsto\:a = 38 + 34$

$\bf\implies \:a \: = \: 72$

$\begin{gathered}\begin{gathered}\bf\: \bf :\longmapsto\:Hence \: - \: \begin{cases} &\bf{a \: = \: 72} \\ \\ &\bf{n \: = \: 17}\\ \\ &\bf{a_n \: = \: 40} \end{cases}\end{gathered}\end{gathered}$