
If the ${{\text{n}}^{{\text{th}}}}$term of an A.P. is ${{\text{t}}_{\text{n}}}$= 3 – 5n, then the sum of the first n terms is:
$
{\text{A}}{\text{. }}\dfrac{{\text{n}}}{2}\left[ {1 - 5{\text{n}}} \right] \\
{\text{B}}{\text{. n}}\left( {1 - 5{\text{n}}} \right) \\
{\text{C}}{\text{. }}\dfrac{{\text{n}}}{2}\left( {{\text{1 + 5n}}} \right) \\
{\text{D}}{\text{. }}\dfrac{{\text{n}}}{2}\left( {1 + {\text{n}}} \right) \\
$
Answer
627.9k+ views
Hint: In order to compute the sum, we use the formula of sum of the first ‘n’ terms of an Arithmetic Mean. To find the unknowns (i.e. a and d) in the formula we calculate a few terms in the progression using the given equation.
Complete step-by-step answer:
Given Data, ${{\text{n}}^{{\text{th}}}}$term of an A.P. is ${{\text{t}}_{\text{n}}}$= 3 – 5n.
Therefore, the first term a = ${{\text{t}}_1}$will be obtained by substituting n = 1 in ${{\text{t}}_{\text{n}}}$= 3 – 5n as follows.
${{\text{t}}_1}$(a) = 3 – (5 × 1) = -2
${{\text{t}}_2}$= 3 – (5 × 2) = -7
${{\text{t}}_3}$= 3 - (5 × 3) = -12
Therefore, d = ${{\text{t}}_2}$- ${{\text{t}}_1}$= -7 – (-2) = -5
We know that the sum of an arithmetic progression with first term ‘a’ and last term${{\text{t}}_{\text{n}}}$, with common difference ‘d’ is:
Sum of n terms of A.P. = $\dfrac{{\text{n}}}{2}\left[ {{\text{2a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right]$
⟹$\dfrac{{\text{n}}}{2}\left[ {{\text{2}}\left( { - 2} \right) + \left( {{\text{n - 1}}} \right) - 5} \right]$
⟹Sum of n terms in A.P. =$\dfrac{{\text{n}}}{2}\left( {1 - {\text{5n}}} \right)$.
Hence Option B is the correct answer.
Note: In order to solve this type of questions the key is to calculate the first few terms of the progression using the given equation, to find out the first term ‘a’ and common difference ‘d’ . Then we apply the formula for the sum of terms in A.P and solve to find the answer.
Complete step-by-step answer:
Given Data, ${{\text{n}}^{{\text{th}}}}$term of an A.P. is ${{\text{t}}_{\text{n}}}$= 3 – 5n.
Therefore, the first term a = ${{\text{t}}_1}$will be obtained by substituting n = 1 in ${{\text{t}}_{\text{n}}}$= 3 – 5n as follows.
${{\text{t}}_1}$(a) = 3 – (5 × 1) = -2
${{\text{t}}_2}$= 3 – (5 × 2) = -7
${{\text{t}}_3}$= 3 - (5 × 3) = -12
Therefore, d = ${{\text{t}}_2}$- ${{\text{t}}_1}$= -7 – (-2) = -5
We know that the sum of an arithmetic progression with first term ‘a’ and last term${{\text{t}}_{\text{n}}}$, with common difference ‘d’ is:
Sum of n terms of A.P. = $\dfrac{{\text{n}}}{2}\left[ {{\text{2a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right]$
⟹$\dfrac{{\text{n}}}{2}\left[ {{\text{2}}\left( { - 2} \right) + \left( {{\text{n - 1}}} \right) - 5} \right]$
⟹Sum of n terms in A.P. =$\dfrac{{\text{n}}}{2}\left( {1 - {\text{5n}}} \right)$.
Hence Option B is the correct answer.
Note: In order to solve this type of questions the key is to calculate the first few terms of the progression using the given equation, to find out the first term ‘a’ and common difference ‘d’ . Then we apply the formula for the sum of terms in A.P and solve to find the answer.
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