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Question: Sum of \({4^{{\text{th }}}}\) and \({8^{{\text{th }}}}\) terms of an AP is 24 and the Sum of \({6^{{...

Sum of 4th {4^{{\text{th }}}} and 8th {8^{{\text{th }}}} terms of an AP is 24 and the Sum of 6th {6^{{\text{th }}}}and 10th {10^{{\text{th }}}}terms of AP is 44. Find the first three terms of AP, also find the sum of 50 terms of AP.

Explanation

Solution

We solve this using the general term of AP which is an=a+(n−1)⋅d{a_n} = a + (n - 1) \cdot d and find the sum using general formula of sum of AP given as: Sn=n2[2a+(n−1)d]{S_n} =\dfrac{n}{2}[2a + (n - 1)d]

Complete step by step solution:
Given that Sum of 4th {4^{{\text{th }}}} and 8th{8^{{\text{th}}}} term is 24.
Now 4th{4^{{\text{th}}}} term is given in general equation by putting n=4, we get,
an=a+(n−1)d⇒a4=a+(4−1)d=a+3d{a_n} = a + (n - 1)d \Rightarrow {a_4} = a + (4 - 1)d = a + 3d
Similarly,8th{8^{{\text{th}}}} term is given in general equation by putting n=8, we get,
a8=a+(8−1)d⇒a8=a+7d{a_8} = a + (8 - 1)d \Rightarrow {a_8} = a + 7d
Now Sum of 4th {4^{{\text{th }}}}and 8th{8^{{\text{th}}}} term is given as,
a4+a8=24⇒a+3d+a+7d=24⇒2a+10d=24{a_4} + {a_8} = 24 \Rightarrow a + 3d + a + 7d = 24 \Rightarrow 2a + 10d = 24
⇒a+5d=12\Rightarrow a + 5d = 12--(A)(A)
Similarly, given that Sum of 6th {6^{{\text{th }}}} and 10th{10^{{\text{th}}}} term is 44.
a6=a+(6−1)d{a_6} = a + (6 - 1)d
⇒a6=a+5d\Rightarrow {a_6} = a + 5d
a10=a+(10−1)d=a+9d{a_{10}} = a + (10 - 1)d = a + 9d
Now, a6+a10=44⇒a+5d+a+9d=44⇒2a+14d=44{a_6} + {a_{10}} = 44 \Rightarrow a + 5d + a + 9d = 44 \Rightarrow 2a + 14d = 44
⇒a+7d=22\Rightarrow a + 7d = 22 --(B)(B)
Solve (A) and (B): (A) - (B)
a+5d−a−7d=12−22a + 5d - a - 7d = 12 - 22
⇒−2d=−10\Rightarrow - 2d = - 10
⇒d=5.\Rightarrow d = 5., Now put value of d in (A)
⇒a=2\Rightarrow a = 2
Now, first three terms of AP given as =a,a+d,a+2d = a,a + d,a + 2d
⇒2,2+5,2+2×5=2,7,10\Rightarrow 2,2 + 5,2 + 2 \times 5 = 2,7,10
First three terms of AP = 2,7,102,7,10
Now, let’s find the sum of first 50 terms of AP
The general formula of sum of series in AP given as:
Sn=n2[2a+(n−1)d]{S_n} =\dfrac{n}{2}[2a + (n - 1)d]
Put n = 50, for the sum of first 50 terms of AP
⇒S50=502[2×2+(50−1)×5]\Rightarrow {S_{50}} =\dfrac{{50}}{2}[2 \times 2 + (50 - 1) \times 5]
⇒S50=25[4+49×5]=25×[4+245]=25×249\Rightarrow {S_{50}} = 25[4 + 49 \times 5] = 25 \times [4 + 245] = 25 \times 249
=6225= 6225
Therefore, Sum of the first 50 terms of AP = 6225.

First three terms of AP = 2,7,102,7,10
Sum of first 50 terms of AP = 6225

Note:
While solving don’t go for the sum first because when you find the value of a (first term) and d(common difference) you can find the sum of terms for that you have to first find out the a and d in this question situation. So, just forget first the 2nd question is also given that is find the sum. Just do questions step by step.