Question

first term of AP is -3 & 10th term is 15, find S10

Answer 1

Ans.
Let the first be a=-3
And 10th term =a+9d (d is common difference)
15=a+9d ( given)
Now, sub. value of a in above equation
15=-3+9d
d=2
Now, S10=n/2[2a+(n-1)d]
S10=10/2[2×-3+(10-1)2]
S10=5[12]
S10=60

Answer 2

The value of S₁₀ = 60

Given :

First term of AP is - 3 & 10th term is 15

To find :

The value of S₁₀

Solution :

Step 1 of 2 :

Write down the first term and 10th term

Here it is given that for the given AP

First term = a = - 3

10th term = 15

Step 2 of 2 :

Find the value of S₁₀

Number of terms = n = 10

S₁₀

$\displaystyle \sf{ = \frac{n}{2}(First \: term + Last \: term ) }$

$\displaystyle \sf{ = \frac{10}{2}(First \: term + 10th \: term ) }$

$\displaystyle \sf{ = 5 \times ( - 3 + 15) }$

$\displaystyle \sf{ = 5 \times 12 }$

$\displaystyle \sf{ = 60}$