Question

In a arithmetic series the sum of 10 numbers is 340 and the sum of first 5 numbers is 95. Then the first number in the series

Answer 1

Solution :-

Given : In a arithmetic series, The sum of 10 numbers is 340 and the sum of first 5 numbers is 95.

10/2 (2a + (10 - 1)d = 340

=> 2a + 9d = 340 × 2/10

=> 2a + 9d = 68

=> 2a = 68 - 9d ______(i)

5/2 (2a + (5 - 1)d = 95

=> (2a + (5 - 1)d = 95 × 2/5

=> 2a + 4d = 38 ______(ii)

Putting the value of 2a in equation (ii) we get,

=> 68 - 9d + 4d = 38

=> - 5d = - 30

=> d = 30/5 = 6

Putting value of d in equation (i),

=> 2a + 9 × 6 = 68

=> 2a = 68 - 54

=> 2a = 14

=> a = 14/2 = 7

Hence,

The first number in the series = 7

Answer 2

Sum of 10 Numbers=340

Sum of AP=n/2(2a+(n-1)d

Where ,

n=Total Number or term

a=First Term of AP

d=Common Difference.

In First Case

Given

Total Term(n)=10

First Term(a)=?

Common Difference (d)=?

Sum=340

$\frac{n}{2} (2a + (n - 1)d = 340 \\ \\ \frac{10}{2} (2a + (10 - 1)d = 340 \\ \\ 5(2a + 9d) = 340 \\ \\ 2a + 9d = \frac{340}{5} \\ \\ 2a + 9d = 68$

In Second Case

Total Term (n)=5

First Term of AP(a)=?

Common Difference (d)=?

Sum=95

$\frac{n}{2} (2a+(n-1)d) = 95 \\ \\ \frac{5}{2} (2a + (5 - 1)d) = 95 \\ \\ 2a + 4d = 95 \times \frac{2}{5} \\ \\ 2a + 4d = 38$

We got

2a+9d=68

2a=68-9d

a=68-9d/2

.And

2a+4d=38

$2( \frac{68 - 9d}{2} ) + 4d = 38 \\ \\ \frac{136}{2} - \frac{18d}{2} + 4d = 38 \\ \\ - \frac{18d}{2} + 4d = 38 - \frac{136}{2} \\ \\ \frac{ - 18d + 8d}{2} = \frac{76 - 136}{2} \\ \\ \frac{ - 10d}{2} = - \frac{ - 60}{2} \\ \\ d = \frac{ - 60}{2} \times \frac{2}{ - 10} \\ \\ d = 6$

We Get Common Difference (d)=6

2a+4d=38

2a+4×6=38

2a+24=38

2a=38-24

2a=14

a=14/2

a=7

$\boxed{\mathtt{\huge{First\:Term(a)=7}}}$