
The first two terms of an A.P are \[{\text{27}}\] and \[{\text{24}}\] respectively. How many terms of the progression are to be added to get -30?
A.15
B.20
C.25
D.18
Answer
593.7k+ views
Hint: Term of an A.P is denoted as \[{{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}\] , so we substitute the given value of \[{\text{27}}\]and \[{\text{24}}\] in the equation and calculate the value of a and d then finally calculate the value of n for \[{{\text{S}}_{\text{n}}}{\text{ = - 30}}\].
Complete step-by-step answer:
As first two term of A.P are 27 and 24 so, equate it to \[{{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}\]
\[
{{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}} \\
\Rightarrow {\text{27 = a and}} \\
\Rightarrow {\text{24 = a + d}} \\
\]
On solving both the above equation it is clear that
\[
{\text{a = 27, substituting its value in 24 = a + d, we get}} \\
{\text{24 = 27 + d}} \\
{\text{d = - 3}} \\
\]
Now , we have to calculate the number of terms up to which sums up to -30
\[
{{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}} \\
{\text{On substituting the value of a, d and }}{{\text{S}}_{\text{n}}}{\text{,we get,}} \\
\Rightarrow {\text{ - 30 = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2(27) + (n - 1)( - 3))}} \\
\Rightarrow {\text{ - 60 = n(57 - 3n)}} \\
{\text{On simplification we get,}} \\
\Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}{\text{ - 57n - 60 = 0}} \\
{\text{On factorisation we get,}} \\
\Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}{\text{ - 60n + 3n - 60 = 0}} \\
\Rightarrow {\text{3n(n - 20) + 3(n - 20) = 0}} \\
\Rightarrow {\text{(3n + 3)(n - 20) = 0}} \\
\Rightarrow {\text{n = 20 or n = - 1}} \\
\]
As value of n cannot be negative,
Hence option (b) is the correct answer.
Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1
Use the data given in the question carefully, place them and form the equation and solve it so that the correct value of required can be obtained.
Complete step-by-step answer:
As first two term of A.P are 27 and 24 so, equate it to \[{{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}}\]
\[
{{\text{T}}_{\text{n}}}{\text{ = a + (n - 1)d}} \\
\Rightarrow {\text{27 = a and}} \\
\Rightarrow {\text{24 = a + d}} \\
\]
On solving both the above equation it is clear that
\[
{\text{a = 27, substituting its value in 24 = a + d, we get}} \\
{\text{24 = 27 + d}} \\
{\text{d = - 3}} \\
\]
Now , we have to calculate the number of terms up to which sums up to -30
\[
{{\text{S}}_{\text{n}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}} \\
{\text{On substituting the value of a, d and }}{{\text{S}}_{\text{n}}}{\text{,we get,}} \\
\Rightarrow {\text{ - 30 = }}\dfrac{{\text{n}}}{{\text{2}}}{\text{(2(27) + (n - 1)( - 3))}} \\
\Rightarrow {\text{ - 60 = n(57 - 3n)}} \\
{\text{On simplification we get,}} \\
\Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}{\text{ - 57n - 60 = 0}} \\
{\text{On factorisation we get,}} \\
\Rightarrow {\text{3}}{{\text{n}}^{\text{2}}}{\text{ - 60n + 3n - 60 = 0}} \\
\Rightarrow {\text{3n(n - 20) + 3(n - 20) = 0}} \\
\Rightarrow {\text{(3n + 3)(n - 20) = 0}} \\
\Rightarrow {\text{n = 20 or n = - 1}} \\
\]
As value of n cannot be negative,
Hence option (b) is the correct answer.
Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1
Use the data given in the question carefully, place them and form the equation and solve it so that the correct value of required can be obtained.
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