Question

The sum of first four terms of an arithmetic sequence is 64 and sum of first 10 terms is 340.what is the sum of first and fourth terms of this sequence?

Answer 1

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

Let assume that

First term of an AP series be a and common difference of an AP series is d.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic progression 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.

According to statement,

$\rm \: S_4 = 64$

$\rm \: \dfrac{4}{2} \bigg(2a + (4 - 1)d \bigg) = 64$

$\rm \:2 \bigg(2a + 3d \bigg) = 64$

$\rm\implies \:2a + 3d = 32 - - - - (1)$

According to statement again

$\rm \: S_{10} = 340$

$\rm \: \dfrac{10}{2} \bigg(2a + (10 - 1)d\bigg) = 340$

$\rm \: 5 \bigg(2a + 9d\bigg) = 340$

$\rm\implies \:2a + 9d = 68 - - - - (2)$

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

$\rm \: 6d = 36$

$\rm\implies \:d = 6$

On substituting the value of d = 6, in equation (1), we get

$\rm \: 2a + 3 \times 6 = 32$

$\rm \: 2a + 18 = 32$

$\rm \: 2a = 32 - 18$

$\rm \: 2a = 14$

$\rm\implies \:a = 7$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic progression 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,

$\rm \: a_1 + a_4$

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

$\rm \: = \:2 a \: + \: 3d$

$\rm \: = \:2 \times 7\: + \: 3 \times 6$

$\rm \: = \: 14 + 18$

$\rm \: = \: 32$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

SHORT CUT TRICK

From equation (1), we have

$\rm \: 2a + 3d = 32$

can be further rewritten as

$\rm \: a + a + 3d = 32$

$\rm\implies \:a_1 + a_4 = 32$

So, Sum of first and fourth term is 32

Answer 2

Answer:

Given :-

• The sum of first four terms of an arithmetic sequence is 64 and sum of first 10 terms is 340.

To Find :-

• What is the sum of first and fourth terms of an AP.

Solution :-

Let,

$\leadsto \bf 1^{st}\: term\: of\: an\: AP =\: a$

$\leadsto \bf 2^{nd}\: term\: of\: an\: AP =\: a + d$

$\leadsto \bf 3^{rd}\: term\: of\: an\: AP =\: a + 2d$

$\leadsto \bf 4^{th}\: term\: of\: an\: AP =\: a + 3d$

According to the question,

$\bigstar$ Sum of first four terms of an AP is 64.

$\implies \sf a + (a + d) + (a + 2d) + (a + 3d) =\: 64$

$\implies \sf a + a + d + a + 2d + a + 3d =\: 64$

$\implies \sf a + a + a + a + d + 2d + 3d =\: 64$

$\implies \sf 4a + 6d =\: 64$

$\implies \sf 2(2a + 3d) =\: 64$

$\implies \sf 2a + 3d =\: \dfrac{64}{2}$

$\implies \sf\bold{\purple{2a + 3d =\: 32\: ------\: (Equation\: No\: 1)}}$

Again,

$\bigstar$ Sum of first 10 terms is 340.

As we know that :

$\longrightarrow \sf\boxed{\bold{\pink{S_n =\: \dfrac{n}{2}\bigg[2a + (n - 1)d\bigg]}}}$

Given :

• n term = 10
• Sum = 340

According to the question by using the formula we get,

$\implies \sf \dfrac{10}{2}\bigg[2a + (10 - 1)d\bigg] =\: 340$

$\implies \sf 5[2a + 9d] =\: 340$

$\implies \sf 10a + 45d =\: 340$

$\implies \sf 5(2a + 9d) =\: 340$

$\implies \sf 2a + 9d =\: \dfrac{340}{5}$

$\implies \sf\bold{\purple{2a + 9d =\: 68\: ------\: (Equation\: No\: 2)}}$

By subtracting both equation (1) and (2) we get,

$\implies \sf 2a + 9d - (2a + 3d) =\: 68 - 32$

$\implies \sf {\cancel{2a}} + 9d {\cancel{- 2a}} - 3d =\: 36$

$\implies \sf 9d - 3d =\: 36$

$\implies \sf 6d =\: 36$

$\implies \sf d =\: \dfrac{36}{6}$

$\implies \sf\bold{\green{d =\: 6}}$

Again, by putting d = 6 in the equation no 2 we get,

$\implies \sf 2a + 9d =\: 68$

$\implies \sf 2a + 9(6) =\: 68$

$\implies \sf 2a + 54 =\: 68$

$\implies \sf 2a =\: 68 - 54$

$\implies \sf 2a =\: 14$

$\implies \sf\bold{\green{a =\: 7}}$

Hence, the required first and fourth terms of an AP are :-

✫ First Term of an AP :-

➸ First Term of an AP = a

➸ First Term of an AP = 7

✫ Fourth Term of an AP :-

➸ Fourth Term of an AP = a + 3d

➸ Fourth Term of an AP = 7 + 3(6)

➸ Fourth Term of an AP = 7 + 18

➸ Fourth Term of an AP = 25

Now, we have to find the sum of first and fourth terms of an AP :-

➵ Sum of first and fourth terms = 7 + 25

➠ Sum of first and fourth terms = 32

$\therefore$ The sum of first and fourth terms of an AP is 32 .

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃