How many terms of the $A.P.$ $27, 24, 21, . . .$ must be taken so that their sum is 105? Which term of the A.P. is zero?
Solution and Explanation
- The given arithmetic progression is 27, 24, 21, ..., with the first term \( a = 27 \) and the common difference \( d = -3 \).
- The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} [2a + (n-1)d] \]
- Substituting the known values:
\[ 105 = \frac{n}{2} [2(27) + (n-1)(-3)] \]
Simplifying:
\[ 105 = \frac{n}{2} [54 - 3n + 3] \] \[ 105 = \frac{n}{2} (57 - 3n) \]
Multiplying both sides by 2:
\[ 210 = n(57 - 3n) \]
Solving the quadratic equation:
\[ 210 = 57n - 3n^2 \] \[ 3n^2 - 57n + 210 = 0 \]
Dividing by 3:
\[ n^2 - 19n + 70 = 0 \]
Solving for \( n \):
\[ n = 7 \text{ or } n = 10 \]
- Therefore, \( n = 7 \) gives the sum as 105.
- To find the term that is zero, we use the formula for the \( n \)-th term:
\[ a_n = a + (n-1)d = 27 + (n-1)(-3) = 0 \]
Solving:
\[ 27 + (n-1)(-3) = 0 \] \[ 27 - 3n + 3 = 0 \] \[ 30 = 3n \] \[ n = 10 \]
So, the term is zero at \( n = 10 \).
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- \(140\)
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Top CBSE X Arithmetic Progression Questions
- In an A.P., if \(a = 8\) and \(a_{10} = -19\), then value of \(d\) is:
- How many terms of the $A.P.$ $27, 24, 21, . . .$ must be taken so that their sum is 105? Which term of the A.P. is zero?
- The sum of the first three terms of an AP is 30 and the sum of the last three terms is 36. If the first term is 9, then the number of terms is :
Assertion (A): The sum of the first fifteen terms of the AP $ 21, 18, 15, 12, \dots $ is zero.
Reason (R): The sum of the first $ n $ terms of an AP with first term $ a $ and common difference $ d $ is given by: $ S_n = \frac{n}{2} \left[ a + (n - 1) d \right]. $- In an A.P., if \(a = 8\) and \(a_{10} = -19\), then value of \(d\) is:
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