What is the answer to this sum 2m^3n^5 / -mn
Answers
Step-by-step explanation:
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✤ Required Answer:
✒ GiveN:
17th term of an AP is 5 more than twice of 8th term.
11th term is 43
✒ To FinD:
Find its 15th term....?
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✤ How to solve?
For the above question, We need to know the formula of nth term of an AP, That is:
\large{ \boxed{ \sf{a_n = a + (n - 1)d}}}
a
n
=a+(n−1)d
Here, an is the last term, a is the first term, n is the no. of terms and d is the common difference. ☃️ So, let's solve this question...
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✤ Solution:
By using the above formula,
a17 = a + 16d
a8 = a + 7d
a11 = a + 10d
According to condition-1)
➝ a17 = 2a8 + 5
➝ a + 16d = 2(a + 7d) + 5
➝ a + 16d = 2a + 14d + 5
➝ a - 2a + 16d - 14d = 5
➝ - a + 2d = 5
➝ - a = 5 - 2d
➝ a = 2d - 5.........(1)
According to condition-2)
➝ a11 = 43
➝ a + 10d = 43
Substituting value of a in this eq.
➝ 2d - 5 + 10d = 43
➝ 12d - 5 = 43
➝ 12d = 48
➝ d = 48/12
➝ d = 4
Putting the value of d in eq.(1),
➝ a = 2(4) - 5
➝ a = 8 - 5
➝ a = 3
We have to find, a15
➝ a15 = a + 14d
➝ a15 = 3 + 14(4)
➝ a15 = 3 + 56
➝ a15 = 59
☀️ 15th term of the AP = 59
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