Answer
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Hint: Consider any AP whose first term is a and the common difference is d. Use the formula that the \[{{n}^{th}}\] term of AP can be written as \[{{a}_{n}}=a+\left( n-1 \right)d\] and the sum of n terms of an AP is \[{{S}_{n}}=\dfrac{n}{2}[2a+\left( n-1 \right)d]\]. Write equations based on the data given in the question and solve them to find the sum of first 15 terms of AP.
Complete step-by-step answer:
We have to find the sum of the first 15 terms of an AP whose fifth term is 26 and ninth term is 42.
We will assume that the first term of an AP is given by a and the common difference is given by d. We will now write any general term of an AP.
So, we can write the\[{{n}^{th}}\] term of AP as \[{{a}_{n}}=a+\left( n-1 \right)d\].
Substituting \[n=5\] in the above equation, we have \[{{a}_{5}}=a+\left( 5-1 \right)d=26\].
Simplifying the above equation, we have \[a+4d=26.....\left( 1 \right)\].
Substituting \[n=9\] in the above formula, we have \[{{a}_{9}}=a+\left( 9-1 \right)d=42\].
Simplifying the above equation, we have \[a+8d=42.....\left( 2 \right)\].
We will now solve equation (1) and (2). Subtracting equation (1) from equation (2), we have \[a+8d-\left( a+4d \right)=42-26\].
Thus, we have \[4d=16\].
\[\Rightarrow d=4\]
So, the value of common difference d is \[d=4\].
Substituting the value \[d=4\] in equation (1), we have \[a+4\left( 4 \right)=26\].
Thus, we have \[a=26-16=10\].
Thus, the first term of AP is \[a=10\].
We will now evaluate the sum of the first 15 terms of AP.
We know that the sum of first n terms of AP is given by \[{{S}_{n}}=\dfrac{n}{2}[2a+\left( n-1 \right)d]\].
Substituting \[n=15,a=10,d=4\] in the above equation, we have \[{{S}_{15}}=\dfrac{15}{2}[2\left( 10 \right)+\left( 15-1 \right)4]\].
Simplifying the above expression, we have \[{{S}_{15}}=570\].
Hence, the sum of the first 15 terms of an AP whose fifth and ninth terms are 26 and 42 respectively is given by 570.
Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. To find the sum of the first 15 terms, we can find all the 15 terms and then add them to get their sum. However, it will be very time consuming. So, it’s better to use the formula of finding sum of n terms to AP.
Complete step-by-step answer:
We have to find the sum of the first 15 terms of an AP whose fifth term is 26 and ninth term is 42.
We will assume that the first term of an AP is given by a and the common difference is given by d. We will now write any general term of an AP.
So, we can write the\[{{n}^{th}}\] term of AP as \[{{a}_{n}}=a+\left( n-1 \right)d\].
Substituting \[n=5\] in the above equation, we have \[{{a}_{5}}=a+\left( 5-1 \right)d=26\].
Simplifying the above equation, we have \[a+4d=26.....\left( 1 \right)\].
Substituting \[n=9\] in the above formula, we have \[{{a}_{9}}=a+\left( 9-1 \right)d=42\].
Simplifying the above equation, we have \[a+8d=42.....\left( 2 \right)\].
We will now solve equation (1) and (2). Subtracting equation (1) from equation (2), we have \[a+8d-\left( a+4d \right)=42-26\].
Thus, we have \[4d=16\].
\[\Rightarrow d=4\]
So, the value of common difference d is \[d=4\].
Substituting the value \[d=4\] in equation (1), we have \[a+4\left( 4 \right)=26\].
Thus, we have \[a=26-16=10\].
Thus, the first term of AP is \[a=10\].
We will now evaluate the sum of the first 15 terms of AP.
We know that the sum of first n terms of AP is given by \[{{S}_{n}}=\dfrac{n}{2}[2a+\left( n-1 \right)d]\].
Substituting \[n=15,a=10,d=4\] in the above equation, we have \[{{S}_{15}}=\dfrac{15}{2}[2\left( 10 \right)+\left( 15-1 \right)4]\].
Simplifying the above expression, we have \[{{S}_{15}}=570\].
Hence, the sum of the first 15 terms of an AP whose fifth and ninth terms are 26 and 42 respectively is given by 570.
Note: Arithmetic Progression is a sequence of numbers such that the difference between any two consecutive terms is a constant. To find the sum of the first 15 terms, we can find all the 15 terms and then add them to get their sum. However, it will be very time consuming. So, it’s better to use the formula of finding sum of n terms to AP.
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