Question

If the third term of an A.P is 7
and the 6th term is 18 then find
the sum of first 5 terms.​

Answer 1

ATQ:

3rd term of an AP = 7

We know that the third term can be written as:

➞ a₃ = 7

We know that a$\sf _{n}$ = a + (n - 1)d.

➞ a + (n - 1)d = 7

➞ a + (3 - 1)d = 7

➞ a + 2d = 7 ⇒ Eq(1)

It has also been given that the 6th term is 18.

We know that the sixth term can be written as:

➞ a₆ = 18

We know that a$\sf _{n}$ = a + (n - 1)d.

➞ a + (n - 1)d = 18

➞ a + (6 - 1)d = 18

➞ a + 5d = 18 ⇒ Eq(2)

Subtracting Eq(1) from Eq(2) we get:

➞ a + 5d - (a + 2d) = 18 - 7

➞ a + 5d - a - 2d = 11

➞ 5d - 2d = 11

➞ 3d = 11

➞ d = 11/3

Substitute the value of 'd' in Eq(1).

➞ a + 2d = 7

➞ a + 2(11/3) = 7

➞ a + 22/3 = 7

➞ a = 7 - (22/3)

➞ a = (21 - 22/3)

➞ a = -1/3

Now, Let's find the sum of the first 5 terms.

$\sf \Longrightarrow S_n = \dfrac{n}{2} \Bigg(2a + (n - 1)d \Bigg)$

Here, n = 5.

$\sf \Longrightarrow S_5 = \dfrac{5}{2} \Bigg(2 \bigg( \dfrac{-1}{3} \bigg) + \bigg(5 - 1 \bigg) \dfrac{11}{3} \Bigg)$

$\sf \Longrightarrow S_5 = \dfrac{5}{2} \Bigg(\dfrac{-2}{3} + \bigg(4\bigg) \dfrac{11}{3} \Bigg)$

$\sf \Longrightarrow S_5 = \dfrac{5}{2} \Bigg(\dfrac{-2}{3} + \dfrac{44}{3} \Bigg)$

$\sf \Longrightarrow S_5 = \dfrac{5}{2} \Bigg(\dfrac{44-2}{3}\Bigg)$

$\sf \Longrightarrow S_5 = \dfrac{5}{2} \Bigg(\dfrac{42}{3}\Bigg)$

$\sf \Longrightarrow S_5 = \dfrac{210}{6}$

$\sf \Longrightarrow S_5 = 35$

Therefore, the sum of the first 5 terms is 35.

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Definition of the terms used:

a ➞ First term of the Arithmetic Progression.

d ➞ Common difference of the Arithmetic Progression.

a$\sf _{n}$ ➞ Position of the nth term.

S$\sf _{n}$ ➞ Sum of the first n terms.

Answer 2

$\bf\huge\underline{{\bigstar ANSWER \bigstar}}$

Given ,

Third Term = 7

Sixth Term = 18

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$\bf\longrightarrow Common \ difference = \dfrac{Term \ difference }{Position \ difference }$

Common difference ,

                         $= \sf \dfrac{18-7}{6-3}$

                         $\bf = \dfrac{11}{3}$

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$\sf 1^{st} \ term = 3^{rd} \ term - 2d$

           $\sf = 7-2\times\dfrac{11}{3}$

           $\sf = 7-\dfrac{22}{3}\\\\= \dfrac{21-22}{3}\\\\= -\dfrac{1}{3}$

$\sf 5^{th} \ term = 1^{st} \ term + 4d$

            $= \sf -\dfrac{1}{3}+4\times\dfrac{11}{3}\\\\= -\dfrac{1}{3}+\dfrac{44}{3}\\\\= \dfrac{44-1}{3}\\\\= \dfrac{43}{3}$  

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$\sf \bf Sum \ of \ first \ n \ terms = \dfrac{n}{2}\times({x_{n}+x_{1})$

$\longrightarrow \sf Sum \ of \ first \ 5 \ terms = \dfrac{5}{2} \times( {\dfrac{43}{3}-\dfrac{1}{3}})$

                                  $\sf = \dfrac{5}{2} \times \dfrac{42}{3}$

                                  $= 5 \times 7 \\\\= 35$

HOPE IT HELPS U ........ :)