Question

the sum of the terms of AP whose 1st term ,last term and common difference are 3,101,and 7 respectively

Answer 1

Given :

• First term ( a ) = 3

• Last term ( An , l ) = 101

• Common difference ( d ) = 7

To Find :

• Sum of the terms in Given A.P

Solution :

First we have to find number of terms

$\large\sf \implies\underline{ \green{A_n = a + (n - 1)d} }\\ \\\sf \implies 101 = 3 + (n - 1)7 \\ \\\sf \implies 98 = (n - 1)7 \\ \\\sf \implies \frac{98}{7} = n - 1 \\ \\\sf \implies n = 14 + 1 \\ \\ \implies \underline{\boxed{ \sf n = 15 }}$

Now sum of terms

$\large\implies \sf \underline{\green{S_n = \frac{n}{2} ( \: a + l \: )}} \\ \\\implies \sf S_{15} = \frac{15}{2} ( \: 3 + 101 \: ) \\ \\\implies \sf S_{15} = \frac{15}{2} ( \: 104 \: ) \\ \\\implies \sf S_{15} =15 \times 52 \\ \\\implies \underline{ \boxed{\sf S_{15} =780}}$