The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
Answers
Answer:
The 26th term is 105.
Step-by-step explanation:
Given :
10th term (a6) = 41
18th term (a17) = 73
an = a + (n – 1) d
a10 = a + (n – 1) d
41 = a + (10 - 1)d
41 = a + 9d
a + 9d = 41
a = 41 - 9d ……….(1)
a18 = a + (n – 1) d
73 = a + (18 - 1) d
73 = a + 17d
73 = (41 - 9d) + 17d
[From eq 1]
73 - 41 = 8d
8d = 32
d = 32/8
d = 4
On substituting the value of d = 4 in (1),
a = 41 - 9d
a = 41 - 9 × 4
a = 41 - 36
a = 5
26th term :
an = a + (n – 1) d
a26 = 5 + (26 - 1) 4
[a = 5 , d = 4, n = 26]
a26 = 5 + 25× 4
a26 = 5 + 100
a26 = 105
26th term = 105
Hence, the 26th term is 105.
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Given :
10th term of an A.P = 41
a + 9d = 41 -----(1)
18th term of an AP = 73
a + 17d = 73 -----(2)
Solve eq - (1) & (2) to find difference (d)
a + 9d = 41
a + 17d = 73
(-)
-------------------
-8d = -32
d = 32/8
d = 4
Difference = 4
Substitute d in eq - (1) to find first term(a)
a + 9(4) = 41
a + 36 = 41
a = 41 - 36
a = 5
First term = 5
26th Term :
= a + 25d
= (5) + 25(4)
= 5 + 100
= 105
Therefore, the 26th term of an A.P is 105