The 5th term of an arithmetic sequence is 20 and the 8th term is 32 . (a) What is the common difference of this sequence? (b) Find its 11th term
Answers
Answer :
Common difference = 4
11th term of A.P = 44
Step-by-step Explanation :
Given : 5th term of A.P = 20
8th term of A.P = 32
To find : a) Common difference = ?
b) 11 th term of A.P = ?
We know that nth term of an AP is given as an = a1 + ( n - 1 ) × d, therefore
A5 = a1 + (5-1) × d
20 = a1 + 4d ----------- (1)
Also,
A8 = a1 + ( 8-1 ) × d
32 = a1 + 7d --------- (2)
Subtract equation (2) by (1)
We get,
20 - 32 = a1 +4d - a1 - 7d
-12 = -3d
d = 12/3 = 4
From equation (1)
a1 = 20 - 4d
a1 = 20 - 4×(4)
a1 = 20 - 16
a1 = 20 - 16 = 4
A11 = 4 + (11-1) × (4)
= 4 + 40
= 44
a) Hence, the common difference is 4
b) Hence, the 11th term is 44
Answer:
The common difference is 4, and 11th term is 44.
Step-by-step explanation:
5th term of the AP = 20
8th term of the AP = 32
Let a be the first term.
nth term of any A.P is,
aₙ = a + (n - 1)d
20 = a + (5 - 1)d
20 - a = 4d
d = (20 - a)/4 ---- (i)
32 = a + (8 - 1)d
32 = a + 7d
32 - a = 7d
d = (32 - a)/7 ---- (ii)
Equate (i) and (ii),
(32 - a)/7 = (20 - a)/4
4(32 - a) = 7(20 - a)
128 - 4a = 140 - 7a
128 - 140 = -7a + 4a
-12 = -3a
a = 4
Put value of 'a' in eq. (i),
d = (20 - 4)/4
d = 16/4
d = 4
Common difference is 4.
11th Term,
a₁₁ = a + (n - 1)d
= 4 + (11 - 1)4
= 4 + 40
= 44
Therefore, the common difference is 4, and 11th term is 44.