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.
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.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
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.
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