Question

find the sum of first 30 terms of an ap 3,7,11​

Answer 1

Answer:

1830

Step-by-step explanation:

$sn = n \div 2(2a + (n - 1)d)$

for the given a.p

a=3,d=4,n=30

so by solving you will get this answer

Answer 2

The sum of first 30 terms of the AP 3 , 7 , 11 is 1830

Given :

The AP 3 , 7 , 11

To find :

The sum of first 30 terms of the AP 3 , 7 , 11

Formula :

Sum of first n terms of an arithmetic progression

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

Where First term = a

Common Difference = d

Solution :

Step 1 of 3 :

Write down the given AP

Here the given AP is 3 , 7 , 11

Step 2 of 3 :

Write down first term and common difference

First term = a = 3

Common Difference = d = 7 - 3 = 4

Step 3 of 3 :

Calculate sum of first 30 terms of the AP 3 , 7 , 11

Number of terms = n = 30

∴ The sum of first 30 terms of the AP 3 , 7 , 11

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

$\displaystyle \sf = \frac{30}{2} \bigg[(2 \times 3) + (30 - 1) \times 4 \bigg]$

$\displaystyle \sf = 15 \times \bigg[6 + 116 \bigg]$

$\displaystyle \sf = 15 \times 122$

$\displaystyle \sf = 1830$

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