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Hint:Given the values of n. Substitute the values in the expression and formulate the series. Prove that the series is in arithmetic progression. So, check that the common difference is the same between two consecutive numbers.

Complete step-by-step answer:

AP is an arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second term minus the first term.

Let us consider ‘d’ as the common difference of the Arithmetic Progression and d is a natural number.

Given the values of n = 1, 2, 3, 4, 5.

Now, substitute these values of n in the expression (3n -10).

When n = 1\[\Rightarrow 3\times 1-10=3-10=-7\].

When n = 2 \[\Rightarrow 3\times 2-10=6-10=-4\].

When n = 3 \[\Rightarrow 3\times 3-10=9-10=-1\].

When n =4 \[\Rightarrow 3\times 4-10=12-10=2\].

When n =5 \[\Rightarrow 3\times 5-10=15-10=5\].

Let us check if -7, -4, -1, 2 and 5 are in AP.

To find if it’s an AP, we should prove that two consecutive terms should have some common difference.

Let us take the 1st and 2nd term.

Common difference = 2nd term - 1st term.

\[\therefore d=-4-\left( -7 \right)=-4+7=3\]

Now check the difference between 2nd term and 3rd term,

\[d=-1-\left( -4 \right)=-1+4=3\]

Check the difference between 3rd term and 4th term.

\[d=2-\left( -1 \right)2+1=3\]

Check the difference is the same for the series.

\[d=5-2=3\]

\[\therefore \]The common difference is the same for the series.

-7, -4, -1, 2, 5

So, the given series is in AP.

Hence proved.

Note:

If we have received different common differences for two consecutive terms, then the series won’t be in Arithmetic progression (AP).

Remember the formula for taking common difference (d). It is the 2nd term minus the 1st term and not the 1st term minus the 2nd term.

Complete step-by-step answer:

AP is an arithmetic progression. It is a sequence of numbers such that the difference between the consecutive terms is constant. Difference here means the second term minus the first term.

Let us consider ‘d’ as the common difference of the Arithmetic Progression and d is a natural number.

Given the values of n = 1, 2, 3, 4, 5.

Now, substitute these values of n in the expression (3n -10).

When n = 1\[\Rightarrow 3\times 1-10=3-10=-7\].

When n = 2 \[\Rightarrow 3\times 2-10=6-10=-4\].

When n = 3 \[\Rightarrow 3\times 3-10=9-10=-1\].

When n =4 \[\Rightarrow 3\times 4-10=12-10=2\].

When n =5 \[\Rightarrow 3\times 5-10=15-10=5\].

Let us check if -7, -4, -1, 2 and 5 are in AP.

To find if it’s an AP, we should prove that two consecutive terms should have some common difference.

Let us take the 1st and 2nd term.

Common difference = 2nd term - 1st term.

\[\therefore d=-4-\left( -7 \right)=-4+7=3\]

Now check the difference between 2nd term and 3rd term,

\[d=-1-\left( -4 \right)=-1+4=3\]

Check the difference between 3rd term and 4th term.

\[d=2-\left( -1 \right)2+1=3\]

Check the difference is the same for the series.

\[d=5-2=3\]

\[\therefore \]The common difference is the same for the series.

-7, -4, -1, 2, 5

So, the given series is in AP.

Hence proved.

Note:

If we have received different common differences for two consecutive terms, then the series won’t be in Arithmetic progression (AP).

Remember the formula for taking common difference (d). It is the 2nd term minus the 1st term and not the 1st term minus the 2nd term.