Question
Find the sum of the first 24 terms of an AP whose nth term is given by tn=3+2n.
Answer: Option D
:
D
Given, tn=3+2n.
∴t1=3+2×1=5
t2=3+2×2=7
t3=3+2×3=9
Note that t2−t1=t3−t2=2.
Thus, the common difference of the AP is 2.
Therefore, the AP is5, 7, 9, 11 . . .
The sum to n terms of an AP with first term a and common difference d is
Sn=n2[2a+(n−1)d].
∴S24=242(2(5)+(24−1)2)
=12(10+46)=672
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:
D
Given, tn=3+2n.
∴t1=3+2×1=5
t2=3+2×2=7
t3=3+2×3=9
Note that t2−t1=t3−t2=2.
Thus, the common difference of the AP is 2.
Therefore, the AP is5, 7, 9, 11 . . .
The sum to n terms of an AP with first term a and common difference d is
Sn=n2[2a+(n−1)d].
∴S24=242(2(5)+(24−1)2)
=12(10+46)=672
Was this answer helpful ?
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