Math, asked by amirkm445, 8 months ago

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. ​

Answers

Answered by gourav667620
3

Answer:

First term ( a ) = ?

Common difference ( d ) = ?

11th term = 16

21st term = 29

Now,

Nth term = First term + ( No. of terms - 1 )Common difference

11th term = a + ( 11 - 1 ) d

16 = a + 10d.

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

Again,

Nth term = First term + ( No. of terms - 1 )Common difference

21st term = a + ( 21 - 1 )d

29 = a + 20d

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

From ( 1 ) and ( 2 ), we get ,

29 - 20d = 16 - 10d

29 - 16 = 20d - 10d

13 = 10d

d = 13 / 10 = 1.3

By substituting the value of 'd' in ( 1 ),

a = 16 - 10d

a = 16 - 10 x 1.3

a = 16 - 13

a = 3.

Now,

Nth term = First term + ( No. of terms - 1 ) Common difference

41st term = 3 + ( 41 - 1 ) 1.3

= 3 + ( 40 ) 1.3

= 3 + 52

= 55.

Answered by anup425337
1

Answer:

Answer:

55

Step-by-step explanation:

In the given A.P. 11th term =16 & 21st term=29

we know that , tn = a+(n-1)d

16 =a+10d ....(I)

similarly, 29= a+20d (II)

subtracting equation I from II

29 = a + 20d

16 = a + 10d

-

------------------------

13 = 10 d

d = 1.3

substituting d = 1.3 in equation I

16 = a+ 10 × 1.3

= a +13

a = 16 - 13

a = 3

now 41th term = 3+(41-1)1.3

= 3 + 52

= 55

please folllw me

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