Problem 1 :
The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Solution :
a17 = a10 + 7
a + 16d = a + 9d + 7
a - a + 16d - 9d = 7
7d = 7
d = 7/7
d = 1
So, the common difference is 1.
Problem 2 :
Which term of the AP :
3, 15, 27, 39,.......... will be 132 more than its 54th term?
Solution :
an = 132 + a54
a + (n - 1) d = 132 + a + 53d
a = 3 d = 15 - 3 = 12
3 + (n - 1) 12 = 132 + 3 + 53 (12)
3 + 12 n - 12 = 135 + 636
-9 + 12 n = 771
12 n = 771 + 9
12 n = 780
n = 780/12
n = 65
Problem 3 :
Two APs have the same common difference. The difference between their 100th term is 100, What is the difference between their 1000th terms?
Solution :
Let the first two terms of two APs a₁ and a₂ respectively and the common difference of these two A.Ps be d
for first A.P
a₁₀₀ = a₁ + (100 - 1) d
= a₁ + 99 d
a₁₀₀₀ = a₁ + (1000 - 1) d
= a₁ + 999 d
For second AP
a₁₀₀ = a₂ + (100 - 1) d
= a₂ + 99 d
a₁₀₀₀ = a₂ + (1000 - 1) d
= a₂ + 999 d
given that,
difference between two 100th term of two APs = 100
(a₁ + 99 d) - (a₂ + 99 d) = 100
a₁ + 99 d - a₂ - 99 d = 100
a₁ - a₂ = 100
Difference between 1000th term of two APs
= (a₁ + 999 d) - (a₂ + 999 d)
= a₁ + 999 d - a₂ - 999 d
= a₁ - a₂
= 100
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