Question

In an A.P If 5th &12th terms are 30 and 65 respectively. what is the sum of first 20 term ?​

Answer 1

Answer:

Step-by-step explanation:

Hence, the sum of first 20 terms for the given A.P. is 1150

Answer 2

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

Let assume that

• First term of an AP = a

• Common difference of an AP = d

Now, it is given that $\rm \: 5^{th}$ term of an AP is 30.

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.

Tʜᴜs,

$\rm \: a_5 = 30$

$\rm \: a + (5 - 1)d = 30$

$\rm\implies \:a + 4d = 30 - - - (1)$

Also, given that

$\rm \: a_{12} = 65$

$\rm \: a + (12 - 1)d = 65$

$\rm\implies \:a + 11d = 65 - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm \: 7d = 35$

$\rm\implies \:d \: = \: 5$

On substituting d = 5, in equation (1), we get

$\rm \: a + 4 \times 5 = 30$

$\rm \: a + 20 = 30$

$\rm\implies \:a \: = \: 10$

Now, we have to find sum of first 20 terms.

We have

$\rm \: a = 10$

$\rm \: d = 5$

$\rm \: n = 20$

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

$\rm \: S_{20}\:=\dfrac{20}{2} \bigg(2 \:(10)\:+\:(20\:-\:1)\:(5)\bigg)$

$\rm \: S_{20}\:= \: 10\bigg(20 + 95\bigg)$

$\rm \: S_{20}\:= \: 10 \times 115$

$\rm\implies \:\boxed{ \rm{ \:\bf \: S_{20}\:= \: 1150 \: \: }}$