Question

find the sum of first 10 terms oof n AP whose first term is 1 and common difference is 3​

Answer 1

Answer:

145

Step-by-step explanation:

S = n/2[2a + (n-1)d]

= 10/2[2×1 + (10-1)3]

= 5[2 + 27]

= 5×29

= 145

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Answer 2

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Given :-

• First term of AP = 1
• Common difference = 3

To find :-

• Sum of first 10 terms.

Solution :-

As we are given the First term, a and the common difference, d we will form an AP using these values till ten terms.

Let's start!

${t_1}$ = 1

Common difference = 3

${t_2}$ = ${t_1}$ + d = 1 + 3 = 4

${t_3}$ = ${t_2}$ + d = 4 + 3 = 7

${t_4}$ = ${t_3}$ + d = 7 + 3 = 10

${t_5}$ = ${t_4}$ + d = 10 + 3 = 13

${t_6}$ = ${t_5}$ + d = 13 + 3 = 16

${t_7}$ = ${t_6}$ + d = 16 + 3 = 19

${t_8}$ = ${t_7}$ + d = 19 + 3 = 22

${t_9}$ = ${t_8}$ + d = 22 + 3 = 25

${t_1_0}$ = ${t_9}$ + d = 25 + 3 = 28

Sequence formed :-

1, 4, 7,10 , 13, 16 , 19, 22, 25, 28.

We have asked to find the sum of first 10 terms of the above AP.

We know that sum of terms in an AP is calculated via the formula,

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

For this AP, we need to calculate $\bf\large{S_1_0}$.

n = 10

a = 1

d = 3

Plug in the values in the formula for the sum of nth an AP's.

$\bf\large{S_1_0}$ = $\bf\large\frac{10}{2}$ [ 2 (1)+ (10 -1) (3) ]

$\bf\large{S_1_0}$ = $\bf\large\frac{10}{2}$ [ 2 + 9 (3) ]

$\bf\large{S_1_0}$ = $\bf\large\frac{10}{2}$ ( 2 + 27 )

$\bf\large{S_1_0}$ = $\bf\large\frac{10}{2}$ (29)

$\bf\large{S_1_0}$ = $\bf\large\frac{10}{2}$ × 29

Dividing 29 by 2,

$\bf\large{S_1_0}$ = 10 × 14.5

$\bf\large{S_1_0}$ = 145

•°• Sum of first ten terms of an AP is 145.