Question

The 11th term and the 21st term of an A.P. are 16 and 29 respectively,then find the 41th term of that A.P.​

Answer 1

Answer:

a11 = 16 , a21 = 29

a + 10d = 16 — 1

a + 20d = 29 — 2

using elimination method

a + 10d = 16

a + 20d = 29

- - -

————————

-10 d = - 13

d = 13/10

put value of d in 1

a + 10×13/10 = 16

a+13 = 16

a = 16-13

a = 3

a41 = = a + 40d

= 3 + 40×13/10

= 3 + 52

= 55

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Answer 2

It is given that a11= 16 and a21 = 29

ᗯE KᑎOᗯ TᕼᗩT,

an = a+ (n-1 ) d

a11 = a + (11-1) d

16 = a+10d ----- (1)

a21 = a + (21-1)d

29 = a+20d -------(2)

(2) - (1)

29 - 16 = 10d

13 = 10d

d = 13/10

Putting the value of d in equation (1),

16 = a + 10 × 13/10

16 = a + 13

a = 3

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41 th term will be

a41 = a + (41-1) d

a41 = 3 + 40 × 13/10

a41 = 3 + (4 × 13)

a41 = 55

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Hope it helps you...