Question

if the 4 th and 7th terms of an A.P are 6 and 18 respectively.Find the sum of first 8 terms of this A.P​

Answer 1

Answer:

64

Explanation:

Given,

4th and 7th term of an A. P. are 6 and 18 respectively.

We need to find its 8th term.

We know,

nth term of an A. P. :-

$\sf \longrightarrow \: a_n = a + (n - 1)d$

Where,

• a denotes the first term.
• n denotes number of term.
• d is the common difference.

So, for the first case :-

Whose 4th term is 6

$\sf \longrightarrow \:a_n = a + (n - 1)d$

$\sf \longrightarrow \:6 = a + (4 - 1)d$

$\sf \longrightarrow \:6 = a + (3)d$

$\sf \longrightarrow \:6 = a + 3d\bf . . . . . . (i)$

And also, for the second case :-

Whose 7th term is 18

$\sf \longrightarrow \:a_n = a + (n - 1)d$

$\sf \longrightarrow \:18 = a + (7 - 1)d$

$\sf \longrightarrow \:18 = a + (6)d$

$\sf \longrightarrow \:18 = a + 6d \bf . . . . . (ii)$

Now, subtracting eq.(i) by (ii)

$\sf \: a + 3d = 6 \: \: \\ {\underline{ \sf{a + 6d = 18}}} \\ \sf \implies \: - 3d = - 12 \\ \sf \implies \: 3d = 12 \\ \sf \implies \: d = \dfrac{12}{3} \\ \sf \therefore \: d = 4$

Substituting ‘d’ in eq.(i) :-

$\sf \implies \: a + 3d = 6$

$\sf \implies \: a + 3(4) = 6$

$\sf \implies \: a + 12 = 6$

$\sf \implies \: a = 6 - 12$

$\sf \implies \: a = - 6$

Now, Sum of first 8th term of the A. P. is given by :-

$\sf \longrightarrow \: \dfrac{n}{2}[2a + (n - 1)d]$

Where,

• n(number of term) = 8
• a(first term) = -6
• d(common difference) = 4

Substituting the values,

$\sf \longrightarrow \: \dfrac{8}{2}[2( - 6) + (8 - 1)(4)]$

$\sf \longrightarrow \: 4[ - 12+ (7)(4)]$

$\sf \longrightarrow \: 4[ - 12+ 28]$

$\sf \longrightarrow \: 4[16]$

$\sf \longrightarrow \: 64$

∴ Sum of first 8th term of the A. P. is 64