Question

the sum of first and 17th term is 40 . sum of first and 18th term is 40
a) find common difference
b) find the sum of 9 and 11th term
c) find the 9th term​

Answer 1

Explanation:

The sum of the 1st and 17th terms of an arithmetic sequence is 40. The sum of its 1st and 18th terms is 43. What is the common difference?

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Nth Term in an Arithmetic progression is given by : Nth term=first term + (n-1)d

Here the sum of ist term and 17th term is 40

Therefore a + a+(16)d = 40

2a + 16d = 40

Also ist term + 18 term =43 (given)

a + a + (17)d =43

2a + 17d = 43

Subtract ist equation from ist

D= 3

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The sum of the 1st and 17th terms of an AP = a +a+16d = 40, or

2a+16d = 40, or

a + 8d = 20 , or

a = 20–8d…(1)

The sum of the 1st and 18th terms of the AP = a +a+17d = 43, or

2a+17d = 43, or

2a = 43–17d …(2)

From (1) and (2)

40–16d = 43–17d, or

17d-16d = 43–40, or

d = 3.

From (1), a = 20 -3d = 20–3*8 = 20–24 = -4.

Answer 2

The answers are  a) 3, b) 37 and c) 20

Given: Sum of first and 17th term = 40

Sum of first and 18th term = 43  [ there might be wrong in given question]

To find: common difference of the series, sum of the 9th and 11th term

and  the 9th term​

Solution: Let a be first term and common difference is d

And nth term of the series $T_{n} = a+(n-1)d$  

⇒ 17th term  $T_{17} = a+(17-1)d$  

⇒ $T_{17} = a+16d$  

⇒ 18th term  $T_{18} = a+(18-1)d$

⇒ $T_{18} = a+17d$

From given Sum of first and 17th term = 40  

⇒ a + a + 16d = 40  

⇒ 2a + 16d = 40 _(1)

Sum of first and 18th term = 43  

⇒ a + a + 17d = 43

⇒ 2a + 17d = 43 _ (2)  

Do (1) - (2) ⇒  2a + 16d - 2a + 17d  = 40 - 43  

⇒ - d = -3   ⇒ d = 3

a) common difference d = 3

Now substitute d = 3 in (1)

⇒  2a + 16(3) = 40  

⇒ 2a + 48 = 40

⇒ 2a = -8   ⇒ a = - 4  

c) 9th term = $T_{9} =$ -4 + (9-1)3

⇒ - 4 + 8 (3) =  -4 + 24 = 20

⇒ 9th term = 20

b) sum of 9th and 11th term

11th term = $T_{11}$ = - 4 + (11-1)3 = -4 + 10(3) = -4 +30 = 26

therefore, $T_{9} +T_{11}$ = 20 +17 = 37  

⇒ sum of 9th and 11th term = 37

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