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
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
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