Question

The sum of first 7 terms of an arithmetic sequence is 119 and the sum of first 11 terms is 275 .find its 4th term?and 6th term ​

Answer 1

Answer:

Given,

Sum of first 7 terms of an AP = 119

Sum of first 17 terms of an AP = 714

Sn = n/2(2a+(n – 1)d]

Case 1,

S7 = 119

7/2[2a+(7 – 1)d] = 119

7/2×2(a+3d) = 119

a+3d = 119/7 = 17

a = 17 – 3d …..(i)

Case 2,

S17 = 714

7/2[2a+(7 – 1)d] = 714

a+8d = 714/17 = 42

By putting (i),

17 – 3d + 8d = 42

So, here we have;

5d = 42 – 17 = 25

d = 25/5 = 5

From Eq.(i) and (ii),

a = 17 – 3 (5) = 17 – 15 = 2

sn = n/2 [2a(n – 1)d]

sn = n/2 [2(2)+(n – 1) 5]

= n/2 [4+5n – 5]

= n/2[5n – 1]

Answer 2

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