Question

The sum of first 7 terms of an arithmetic sequence is 119 and the sum of first 20 terms is860.
a)What is the 4th term?
b)What is its 17th term ?

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Answer 1

Answer:-

Given:

Sum of first 7 terms of an AP (S7) = 119

Sum of first 20 terms of the AP (S20) = 860.

We know that,

Sum of first n terms of an AP = n/2 * [ 2a + (n - 1)d ]

Hence,

→ S(7) = 7/2 * [ 2a + (7 - 1)d ]

→ S(7) = 7/2 * [ 2a + 6d ]

→ 7/2 * [ 2 ( a + 3d) ] = 119

→ 7 ( a + 3d ) = 119

→ a + 3d = 17

We know,

nth term of an AP (an) = a + (n - 1)d

Hence,

a(4) = a + (4 - 1)d

→ a(4) = a + 3d

→ a(4) = 17 - (a)

Similarly,

S(20) = 20/2 * [ 2a + (20 - 1)d ]

→ 860 = 10 (2a + 19d)

→ 860/10 = 2a + 19d

→ 2a + 19d = 86

→ (a + 3d) + a + 16d = 86

→ (17) + (a + 16d) = 86

→ a + 16d = 86 - 17

→ a(17) = 69 - (b)

[ a(17) = a + (17 - 1)d ]

Hence, the 4th and 17th terms of the given AP are 17 , 69.

Answer 2

Answer:

(a). 17, (b). 69

Step-by-step explanation:

$S_{n}$ = $\frac{n}{2}$(2$a_{1}$ + (n - 1)d)

$a_{n}$ = $a_{1}$ + (n - 1)d

~~~~~~~~~~~~~

$\frac{7}{2}$ (2$a_{1}$ + 6d) = 119

$\frac{10}{2}$ (2$a_{1}$ + 19d) = 860

2$a_{1}$ + 19d = 86 .... (1)

2$a_{1}$ + 6d = 34 ..... (2)

(1) - (2)

19d - 6 d = 86 - 34

13d = 52 ===> d = 4

2$a_{1}$ + 24 = 34 ===> $a_{1}$ = 5

$a_{4}$ = 5 + (4 - 1)×4 = 17  

$a_{17}$ = 5 + (17 - 1)×4 = 69