Question

an ap is formed by the multiples of 4 between 50 and 350. find the sum of the first twelve terms of the ap​

Answer 1

Answer :-

Sum of 12 terms is 888.

Note :-

• Numbers are said to be in AP , if a constant number is added to form the next term.

• The nth term of an AP is given by a + (n-1)d.

• Sum of n terms of a AP is given by n/2 [ 2a + (n-1)d]

• Arthemetic mean of two numbers is given by (a+b)/2 .

Explanation :-

Given that , an AP is formed by the multiples of 4 between 50 and 350. find the sum of the first twelve terms of the AP . So ;

• First multiple of 4 b/w 50 & 350=52
• Last multipe of 4 b/w 50 & 350= 348.

We know ,the formula to find the sum of n terms of AP as ;

$=> \red{S_n =\dfrac{n}{2}[2a+(n-1)d]}\\\\=> S_{12} =\dfrac{12}{2}[2\times 52 + ( 12-1)4\\\\=> S_{12} = 6 [ 104 + 11\times 4 ] \\\\=> S_{12} = 6 [ 104 + 44]\\\\=> S_{12} = 6 \times 148 \\\\=>\sf \pink{S_{12} = 888 }$

Answer 2

Answer:

888

Step-by-step explanation:

Here the first term is 52

52+56+60+........

since we need sum of first twelve terms only

4(13+14+15+..+24). (By taking 4 common)

In the simplied series a=13,d=1, last =25,n=12

s=n/2(first+last)

s=12/2(13+24)

s=222

s=4×222=888