Math, asked by GGOOUURRAAVV, 11 months ago

T7=13,S14=203 then find S8​

Answers

Answered by knjroopa
10

Answer:

44

Step-by-step explanation:

Given  

T7 = 13, S14 = 203 then find S8

ANSWER

Given in an A.P T7 = 13  

We know that nth term of an A.P is  

 T n = a + (n – 1) d

T 7 = a + (7 – 1) d  

13 = a + 6d ------------(1)

Now S 14 = 203

We know that

Sum to n terms of an A.P is given by

S n = n/2 (2 a + (n – 1)d)

S 14 = 14/2 (2a + (14 – 1)d)

203 = 7(2a + 13d)

 203/7 = 2a + 13d

 2a + 13d = 29------------(2)

From 1 and 2 we get

a  + 6d = 13

2a + 13d = 29

Multiplying equation 1 by 2 and subtracting we get

2a + 12d = 26

2a + 13d = 29

d = 3

Substituting in equation 1 we get

a + 6(3) = 13

a = 13 – 18

a = - 5

Now we need to find sum to n terms

Sn = n/2 (2a + (n – 1)d)

S8 = 8/2 (2(-5) + (8 – 1)3

S8 = 4(-10 + 21)

S8 = 44

Answered by skmittal15
1

Answer:

44

Step-by-step explanation:

Given

T7 = 13, S14 = 203 then find S8

ANSWER

Given in an A.P T7 = 13

We know that nth term of an A.P is

T n = a + (n – 1) d

T 7 = a + (7 – 1) d

13 = a + 6d ------------(1)

Now S 14 = 203

We know that

Sum to n terms of an A.P is given by

S n = n/2 (2 a + (n – 1)d)

S 14 = 14/2 (2a + (14 – 1)d)

203 = 7(2a + 13d)

203/7 = 2a + 13d

2a + 13d = 29------------(2)

From 1 and 2 we get

a + 6d = 13

2a + 13d = 29

Multiplying equation 1 by 2 and subtracting we get

2a + 12d = 26

2a + 13d = 29

d = 3

Substituting in equation 1 we get

a + 6(3) = 13

a = 13 – 18

a = - 5

Now we need to find sum to n terms

Sn = n/2 (2a + (n – 1)d)

S8 = 8/2 (2(-5) + (8 – 1)3

S8 = 4(-10 + 21)

S8 = 44

❤️THANKS AND MARK AS BRAINYLIST

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