Question

Find the sum of the first 10 positive integers divisible by 2.​

Answer 1

Step-by-step explanation:

A.P= 2,4,6,........

n=10

d=2

a=2

Sn=

$\frac{n}{2} (2a + (n - 1)d) \\ \frac{10}{2} (2 \times 2 + (10 - 1)2) \\ 5(4 + 9 \times 2) \\ 5(4 + 18) \\ 5 \times 22 \\ 110$

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Answer 2

$\huge \underline{ \underline{ \bf \purple{Solution :}}}$

First 10 integers which is divisible by 2 are :-

→ 2 , 4 , 6 .....

Here

• First term ( a ) = 2

• Common difference ( d ) = 2

• Number of terms ( n ) = 10

Now , By using A.P formula

$\implies \underline{ \boxed{\sf \green{S_n = \frac{n}{2} \bigg [2a + (n - 1)d \bigg]}}} \\ \\\implies \sf S_n = \frac{10}{2} \bigg [2 \times 2 + (10 - 1)2 \bigg] \\ \\\implies \sf S_n =5\bigg [4+ 18 \bigg] \\ \\\implies \sf S_n =5 \times 22 \\ \\ \implies \underline{ \boxed{\sf \orange{ S_n =110}}}$