Math, asked by sriyadutta2464, 11 months ago

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


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Answers

Answered by Anonymous
119

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

Answered by pratyush4211
51

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}}}


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