# TThe \[{{4}^{th}}\],\[{{42}^{nd}}\] and last term of an AP are 0, -95 and -125 respectively. Find the first term and the number of terms.

Answer

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Hint: Consider any AP whose first term is ‘a’ and the common difference is ‘d’. Write \[{{n}^{th}}\] term of AP as \[{{a}_{n}}=a+\left( n-1 \right)d\]. Write equations for terms of an AP and solve them using elimination method to get the value of a, n and d.

Complete step by step answer:

We have an AP whose \[{{4}^{th}},{{42}^{nd}}\] and last term is 0, -95 and -125. We have to find the first term of AP and the number of terms in AP.

Let’s assume that the first term of AP is ‘a’ and the common difference is ‘d’.

We know that we can write the \[{{n}^{th}}\] term of AP as \[{{a}_{n}}=a+\left( n-1 \right)d\].

Substituting \[n=4\] in the above equation, we have \[{{a}_{4}}=a+\left( 4-1 \right)d\]. Thus, we have \[a+3d=0.....\left( 1 \right)\].

Substituting \[n=42\] in the above equation, we have \[{{a}_{42}}=a+\left( 42-1 \right)d\]. Thus, we have \[a+41d=-95.....\left( 2 \right)\].

Subtracting equation (1) from equation (2), we have \[a+41d-\left( a+3d \right)=-95-0\].

Thus, we have \[38d=-95\Rightarrow d=-\dfrac{95}{38}=-2.5\].

Substituting the value \[d=-2.5\] in equation (1), we have \[a+3\left( -2.5 \right)=0\].

Thus, we have \[a=-7.5\].

We know that the last term of this AP is -125. Let’s assume that there are ‘x’ terms in this AP.

Thus, we have \[{{a}_{x}}=a+\left( x-1 \right)d=-125\].

Substituting \[a=-7.5,d=-2.5\] in the above formula, we have \[-7.5+\left( x-1 \right)\left( -2.5 \right)=-125\].

Simplifying the above equation, we have \[\left( x-1 \right)\left( -2.5 \right)=-117.5\].

Thus, we have \[x-1=\dfrac{-117.5}{-2.5}=47\].

So, we have \[x=47+1=48\].

Hence, the first term of this AP is -7.5 and the number of terms is 48.

Note: One must clearly know the definition of AP. Arithmetic Progression (AP) is the sequence of numbers in which the difference of two consecutive terms is a constant. We can also solve these linear equations by substitution method. We can check if the calculated solutions are correct or not by substituting the values in the equations and checking if they satisfy the equations or not.

Complete step by step answer:

We have an AP whose \[{{4}^{th}},{{42}^{nd}}\] and last term is 0, -95 and -125. We have to find the first term of AP and the number of terms in AP.

Let’s assume that the first term of AP is ‘a’ and the common difference is ‘d’.

We know that we can write the \[{{n}^{th}}\] term of AP as \[{{a}_{n}}=a+\left( n-1 \right)d\].

Substituting \[n=4\] in the above equation, we have \[{{a}_{4}}=a+\left( 4-1 \right)d\]. Thus, we have \[a+3d=0.....\left( 1 \right)\].

Substituting \[n=42\] in the above equation, we have \[{{a}_{42}}=a+\left( 42-1 \right)d\]. Thus, we have \[a+41d=-95.....\left( 2 \right)\].

Subtracting equation (1) from equation (2), we have \[a+41d-\left( a+3d \right)=-95-0\].

Thus, we have \[38d=-95\Rightarrow d=-\dfrac{95}{38}=-2.5\].

Substituting the value \[d=-2.5\] in equation (1), we have \[a+3\left( -2.5 \right)=0\].

Thus, we have \[a=-7.5\].

We know that the last term of this AP is -125. Let’s assume that there are ‘x’ terms in this AP.

Thus, we have \[{{a}_{x}}=a+\left( x-1 \right)d=-125\].

Substituting \[a=-7.5,d=-2.5\] in the above formula, we have \[-7.5+\left( x-1 \right)\left( -2.5 \right)=-125\].

Simplifying the above equation, we have \[\left( x-1 \right)\left( -2.5 \right)=-117.5\].

Thus, we have \[x-1=\dfrac{-117.5}{-2.5}=47\].

So, we have \[x=47+1=48\].

Hence, the first term of this AP is -7.5 and the number of terms is 48.

Note: One must clearly know the definition of AP. Arithmetic Progression (AP) is the sequence of numbers in which the difference of two consecutive terms is a constant. We can also solve these linear equations by substitution method. We can check if the calculated solutions are correct or not by substituting the values in the equations and checking if they satisfy the equations or not.

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