Question

*.The sum of first n terms of an AP whose first term is a and common difference is d, is* 1️⃣ n/2 [a+(n-1)d] 2️⃣ n/2[2a+(n-1)d] 3️⃣ n/2[a+(n-1)d] 4️⃣ none
only give right ans ​

Answer 1

Answer:

n/2{2a+(n-1)d} is the answer of these question.

Answer 2

Solution :-

we know that,

• A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.

• The nth term of an AP is given as :-

• T(n) = a + (n-1)•d , where a is the first term and d is the common difference.

• The common difference of an AP is given as :-

• d = T(n) - T(n-1)

• The sum up to nth terms of an AP is given as :-

• S(n) = (n/2)[2a + (n - 1)d]

• The nth term of an AP is also given as :-

• T(n) = S(n) - S(n-1)

given that,

→ First term = a

→ common difference = d

so,

→ nth term = a + (n - 1)d .

then,

→ Sn = (n/2)[first term + Last term]

→ Sn = (n/2)[a + a + (n - 1)d]

→ Sn = (n/2)[2a + (n - 1)d] Option (2) (Ans.)

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