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Find the 68th term of the following arithmetic sequence.
4, 10, 16, 22, …

Find the 68th term of the following arithmetic sequence. 4, 10, 16, 22, …-example-1
User Vikram R
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2 Answers

15 votes
15 votes

The given Arithmetic Sequence is 4 , 10 , 16 , 22 , ... and we need to find the 68th term of the sequence , but let's recall that for any sequence in AP , the nth term is :


  • {\boxed{\bf{a_(n)=a+(n-1)d}}}

Where ,
\bf a is the first term of the sequence and
\bf d is the common difference (
{\bf{a_(n)-a_(n-1)}} ) and
\bf a_n is nth term. So , now in this question , a = 4 and d = 10 - 4 = 6 . Now , rhe 68th term will be :


{:\implies \quad \sf a_(68)=4+(68-1)6}


{:\implies \quad \sf a_(68)=4+(67)6}


{:\implies \quad \sf a_(68)=4+402=406}

Hence , the 68th term is 406

User MegaCookie
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\bold{\huge{\underline{ Solution }}}

Given :-

  • Here, We have given the arithmetic sequence that is 4 , 10 , 16 , 22 ...and so on

To Find :-

  • We have to find the 68th term of the given AP

Let's Begin :-

Here, we have

  • Arithmetic sequence :- 4 , 10 , 16 , 22

We have to determine the 68th term of given AP

Therefore,

By using an formula that is,


\bold{\red{ an = a1 + (n - 1)d }}

  • Here, a1 = first term
  • n = number of terms
  • d = common difference
  • an = term number

For finding common difference of AP

  • Subtract preceeding term from succeeding term
  • That is, a2 - a1

Here, common difference will be


\sf{ = 10 - 4 }


\sf{ = 6 }

Thus, The common difference of the given AP is 6

Now, Subsitute the given values in the above an formula :-


\sf{ an = 4 + (68 - 1)6 }


\sf{ an = 4 + 67 {*} 6 }


\sf{ an = 4 + 402 }


\sf{ an = 406 }

Hence, The 68th term of the given AP is 406

Some basic details about AP

  • Arithmetic progression is the sequence of numbers that have same common difference between each succeeding and preceeding term.
  • For finding terms,

  • \bold{\red{ an = a1 + (n - 1)d }}
  • For finding sum of terms

  • \bold{\red{ sn = }}{\bold{\red{(n)/(2)}}}{\bold{\red{ [2a + ( n - 1)d ]}}}
User Ram Chander
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