How many terms are there in the A.P : −1, 6−5,−32 , 2−1....310 ?

Here, first term a=−1 and the common difference d= 6−5−(−1)=61.Let n be number of terms of given AP.Then, an=310⇒a+(n−1)d=310⇒−1+(n−1)×61=310⇒(n−1)×61=313⇒n−1=26⇒n=27Therefore, there are 27 terms.

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How many terms are there in the A.P : -1, {-5}{6},-{2}{3} , {

Question

How many terms are there in the A.P : $- 1$ , $\frac{- 5}{6}$ , $- \frac{2}{3}$ , $\frac{- 1}{2}$ .... $\frac{10}{3}$ ?

27

18

16

9

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Evelyn Discussed

How many terms are there in the A.P :

- 1

\frac{- 5}{6}

- \frac{2}{3}

\frac{- 1}{2}

....

\frac{10}{3}

6 mins ago

Discuss this question LIVE

6 mins ago

Text solutionVerified

Here, first term

a = - 1

and the common difference

d = \frac{- 5}{6} - (- 1) = \frac{1}{6} .

Let

n

be number of terms of given AP.
Then,

a_{n} = \frac{10}{3}

\Rightarrow a + (n - 1) d = \frac{10}{3}

\Rightarrow - 1 + (n - 1) \times \frac{1}{6} = \frac{10}{3}

\Rightarrow (n - 1) \times \frac{1}{6} = \frac{13}{3}

\Rightarrow n - 1 = 26

\Rightarrow n = 27

Therefore, there are

27

terms.

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Question Text	How many terms are there in the A.P : $- 1$ , $\frac{- 5}{6}$ , $- \frac{2}{3}$ , $\frac{- 1}{2}$ .... $\frac{10}{3}$ ?
Answer Type	Text solution:1
Upvotes	150

How many terms are there in the A.P : −1, 6−5​,−32​ , 2−1​....310​ ?

Text solutionVerified

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How many terms are there in the A.P : $- 1$ , $\frac{- 5}{6}$ , $- \frac{2}{3}$ , $\frac{- 1}{2}$ .... $\frac{10}{3}$ ?