Question

Find the sum of first 15 terms of the AP : 3 , 8 , 13 , 18 , 23

Answer 1

Given : AP is 3, 8, 13, 18, 23 . . .

To find : Sum of its first 15 terms

Solution :

AP ( Arithmetic Progression ) is a sequence of numbers in which common difference between two consecutive terms is always same throughout every consecutive terms.

The general form of sum of n terms of AP is :-

• Sn = n/2 [ 2a + (n - 1) d ]

Here,

• Sn = Sum of n terms of AP
• a = First term of AP
• d = Common difference of AP
• n = number of terms of AP

Common difference = a2 - a1

Common difference = 8 - 3

Common difference = 5

If we substitute n = 15 in the formula of sum of AP, we will get sum of it's 15 terms. Also substitute value of a and d.

→ S15 = 15/2 [ 2(3) + ( 15 - 1 )( 5 ) ]

→ S15 = 15/2 [ 6 + (14) (5) ]

→ S15 = 15/2 [ 6 + 70 ]

→ S15 = 15/2 [ 76 ]

→ S15 = 15 × 38

→ S15 = 570

Hence the sum of 15 terms of AP is 570.

Answer 2

SOLUTION

TO DETERMINE

The sum of first 15 terms of the AP : 3 , 8 , 13 , 18 , 23

CONCEPT TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

Where First term = a

Common Difference = d

EVALUATION

Here the given arithmetic progression is

3 , 8 , 13 , 18 , 23

First term = a = 3

Common Difference = d = 8 - 3 = 5

Number of terms = n = 15

Hence the required sum

$\displaystyle \sf = \frac{n}{2} \bigg[2a + (n - 1)d \bigg]$

$\displaystyle \sf = \frac{15}{2} \times \bigg[(2 \times 3)+ 5 \times (15- 1) \bigg]$

$\displaystyle \sf = \frac{15}{2} \times \bigg[6+ (5 \times 14) \bigg]$

$\displaystyle \sf = \frac{15}{2} \times \bigg[6+ 70 \bigg]$

$\displaystyle \sf = \frac{15}{2} \times 76$

$\displaystyle \sf = 15 \times 38$

$\displaystyle \sf = 570$

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