Question

if the sum of fist 6 terms of ap is 96 and sum of first 10 term is 240 ,then find the sum of 20 term of ap​

Answer 1

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

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.

Tʜᴜs,

According to statement,

Given that

☆ Sum of first 6 terms of an AP is 96.

$\rm :\longmapsto\:S_6 = 96$

$\rm :\longmapsto\:\dfrac{6}{2} \bigg(2 \:a\:+\:(6\:-\:1)\:d \bigg) = 96$

$\rm :\longmapsto\:2a + 5d = 32 - - - (1)$

Also,

Given that

☆ Sum of 10 terms is 240

$\rm :\longmapsto\:S_{10} = 240$

$\rm :\longmapsto\:\dfrac{10}{2} \bigg(2 \:a\:+\:(10\:-\:1)\:d \bigg) = 240$

$\rm :\longmapsto\:2a + 9d = 48 - - - (2)$

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

$\rm :\longmapsto\:4d = 16$

$\bf\implies \:d = 4$

○ On Substituting d = 4, in equation (1), we get

$\rm :\longmapsto\:2a + 5 \times 4 = 32$

$\rm :\longmapsto\:2a + 20 = 32$

$\rm :\longmapsto\:2a = 32 - 20$

$\rm :\longmapsto\:2a = 12$

$\bf\implies \:a = 6$

So,

Now we have ,

• First term of AP, a = 6

• Common difference of AP, d = 4

• Number of terms, n = 20

Sum of 20 terms, is

$\rm :\longmapsto\:S_{20}\:=\dfrac{20}{2} \bigg(2 \times 6\:+\:(20\:-\:1)\:4 \bigg)$

$\rm \: \: = \: \: 10(12 + 76)$

$\rm \: \: = \: \: 10(88)$

$\rm \: \: = \: \: 880$

$\bf\implies \:S_{20} = 880$

Additional Information :-

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.