Question

find the sum of first 23 terms of the arithmetic progression 2,7,12​

Answer 1

Answer:

hey mate here is the answer

Step-by-step explanation:

ap = n/2= 2a+(n-1)d

n = 10, a= 2 d= 2

10/2= (2×2 +[10-1] ×5)

5(3+9×5)

5(4+45)

= 5×49

= 245

ans - sum of first 23rd term of ap 2,7,12 is 245

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Answer 2

Answer:

The sum of first 23 terms of the arithmetic progression is 1311

Step-by-step explanation:

Given: The arithmetic progression 2,7,12​

To find: The sum of first 23 terms of the arithmetic progression

Formula used: $S_{n} = \frac{n}{2} [2a +(n-1)d$

Solution:

Arithmetic progression:

An arithmetic progression or sequence is a set of numbers where, for every pair of consecutive terms, the second number is derived by adding a predetermined number to the first.

The arithmetic progression term is 2,7,12

We know that ,

Sum of n terms is $S_{n} = \frac{n}{2} [2a +(n-1)d$

a = first term

d = common difference

n =  number of terms

From the given arithmetic progression 2,7,12

a = 2

d = 7-2 = 5

n = 23

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

$S_{23} = \frac{23}{2} [2(2) +(23-1)5]$

$S_{23} = \frac{23}{2} [4+(23-1)5]$

$S_{23} = \frac{23}{2} [4 +(22)5]$

$S_{23} = \frac{23}{2} [4 +110]$

$S_{23} = \frac{23}{2} [114]$

$S_{23} = 23(57)$

$S_{23} = 1311$

Final answer:

The sum of first 23 terms of the arithmetic progression is 1311

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