Question

1. Find the sum of 14 terms of an A.P. whose nth term is given by an = 3n+ 5
a. 385
b. 375
C. 395
d. 365



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Answer 1

We've been given that the nth term of this AP is given by 3n + 5.

We can find the other terms of the AP by taking different values for 'n', then find the sum of the first 14 terms with the help of the first term and common difference.

Let's say n = 1.

$\sf \Longrightarrow a_{n} = 3n+ 5$

$\sf \Longrightarrow a_{1} = 3\{1\}+ 5$

$\sf \Longrightarrow a_{1} = 3 + 5$

$\sf \Longrightarrow a_{1} = 8$

Let's say n = 2.

$\sf \Longrightarrow a_{n} = 3n+ 5$

$\sf \Longrightarrow a_{2} = 3\{2\}+ 5$

$\sf \Longrightarrow a_{2} = 6 + 5$

$\sf \Longrightarrow a_{2} = 11$

Let's say n = 3.

$\sf \Longrightarrow a_{n} = 3n+ 5$

$\sf \Longrightarrow a_{3} = 3\{3\}+ 5$

$\sf \Longrightarrow a_{3} = 9 + 5$

$\sf \Longrightarrow a_{3} = 14$

Therefore:

First term (a) = 8

Common difference (d) = 14 - 11 = 3

Now, we'll find the sum of the first 14th terms.

a = 8

d = 3

n = 14

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

$\sf \Longrightarrow S_{14} = \dfrac{14}{2}\Big(2(8) + (14 - 1)3 \Big)$

$\sf \Longrightarrow S_{14} = 7\Big(16 + (13)3\Big)$

$\sf \Longrightarrow S_{14} = 7\Big(16 + 39 \Big)$

$\sf \Longrightarrow S_{14} = 7\Big(55\Big)$

$\sf \Longrightarrow S_{14} = 385$

Answer: Option(a) 385

Answer 2

★Question:–

Find the sum of 14 terms of an A.P. whose nth term is given by an = 3n+ 5

a. 385

b. 375

C. 395

d. 365.

★Answer:–

option (a) 385.

★Solution:–

given

$a_n = 3n + 5$

let n=1 in given equation

$a_1 = 3 + 5 \\ a_1 = 8$

let n=2

$a_2 = 6 + 5 \\ a_2 = 11$

let n=3

$a_3 = 9 + 5 \\ a_3 = 14$

therefore,

AP is 8,11,14.

→ a=8

$d = a_2 - a_1 = 11 - 8 = 3$

→d=3

→n=14

★Formula used:–

$S_n = \frac{n}{2} (2a + (n - 1)d)$

$S_1_4 = \: \frac{14}{2} (2(8) + (13)3)$

$S_1_4 = 7(16 + (13)3)$

$S_1_4 = 7(16 + 39)$

$S_1_4 = 7 \times 55$

$S_1_4 = 385$

therefore,

Sum of 14 term of AP is 385.