Question

The sum of the first 30 terms of an A.P. is 1920. If the fourth term is 18,
find its 11th term.​

Answer 1

Answer:

S30 = 30/2(2a+(n-1)d) = 1920

= 15(2a+(n-1)d = 1920

2a+(n-1)d = 1920/15= 128

2a+29d = 128 ...........(1)

a+3d = 18 .........(2) (multiply by 2)

2a+6d = 36...... (subtracting from eq 1)

we get.

23d = 128-36 = 92

d = 92/23 = 4

then

a = 18-3d = 18-3×4 = 18-12 = 6

then

11th term = a+10d = 6+10×4 = 46

hoope it helps....

Answer 2

★ Given :

• Sum of first 30 terms of the A.P is 1920.

• Fourth term of the A.P is 18

$\rule{100}{2}$

★ To Find :

• We have to find the 11th term of the A.P

$\rule{100}{2}$

★ Solution :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{S_n = \frac{n}{2} \bigg(2a + (n - 1)d \bigg)}}}}}$

Putting Values

$\sf{\dashrightarrow 1920 = \frac{30}{2} \bigg(2a + (30 - 1)d \bigg)} \\ \\ \sf{\dashrightarrow 1920 = 15 (2a + 29d)} \\ \\ \sf{\dashrightarrow \frac{1920}{15} = 2a + 29d} \\ \\ \sf{\dashrightarrow 2a + 29d = 128 .....(1)}$

$\rule{150}{2}$

$\sf{\dashrightarrow A_4 = a + 3d} \\ \\ \sf{\dashrightarrow a + 3d = 18} \\ \\ \sf{\dashrightarrow a = 18 - 3d .......(2)}$

➳ Using equation (2), put value of a in equation (1).

$\sf{\dashrightarrow 2(18 - 3d) + 29d = 128} \\ \\ \sf{\dashrightarrow 36 - 6d + 29d = 128} \\ \\ \sf{\dashrightarrow 23d = 128 - 36} \\ \\ \sf{\dashrightarrow 23d = 92} \\ \\ \sf{\dashrightarrow d = \frac{92}{23}} \\ \\ \sf{\dashrightarrow d = 4}$

$\rule{100}{2}$

➳ Put value of d in equation (2).

$\sf{\dashrightarrow a = 18 - 3(4)} \\ \\ \sf{\dashrightarrow a = 18 - 12} \\ \\ \sf{\dashrightarrow a = 6}$

$\rule{150}{2}$

Now,

We know that,

$\Large{\implies{\boxed{\boxed{\sf{A_{11} = a + 10d}}}}}$

$\sf{\dashrightarrow A_{11} = 6 + 10(4)} \\ \\ \sf{\dashrightarrow A_{11} = 6 + 40} \\ \\ \sf{\dashrightarrow A_{11} = 46}$

$\underline{\sf{\therefore \: 11^{th} \: term \: of \: the \: A.P \: is \: 46.}}$