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In an AP, (a=14), (d=7), and (n=28). Find the sum of the first (28) terms.

Answer and explanation

Correct answer: (3038)

Direct answer: A, 3038. For an arithmetic progression, the sum of the first n terms is \(S_n=\frac{n}{2}[2a+(n-1)d]\). Here \(a=14\), \(d=7\), and \(n=28\). First calculate \(n-1=27\). Then \((n-1)d=27\times7=189\). Next, \(2a=2\times14=28\), so the bracket is \(28+189=217\). Finally, \(\frac{28}{2}\times217=14\times217=3038\). Therefore option A is correct. Option B, 3024, does not result from the correct substitution and may come from a small arithmetic or term-counting error. Option C, 3052, is also not the value of the formula. Option D, 3010, is likewise incorrect. Another check uses the last term: \(l=a+(n-1)d=14+189=203\), so \(S_{28}=\frac{28}{2}(14+203)=14\times217=3038\). Memory cue: never use \(nd\); the formula needs \((n-1)d\).

Related tags

Arithmetic ProgressionAp SumHard

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