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If the sum of 4 terms of an A.P. is 36, ...

If the sum of 4 terms of an A.P. is 36, then find the A.M of the A.P.

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To solve the problem step by step, we will follow the logic presented in the video transcript. ### Step 1: Understand the Problem We are given that the sum of 4 terms of an Arithmetic Progression (A.P.) is 36. We need to find the Arithmetic Mean (A.M.) of these terms. ### Step 2: Represent the 4 Terms of the A.P. Let the four terms of the A.P. be represented as: - First term: \( a - 3d \) - Second term: \( a - d \) - Third term: \( a + d \) - Fourth term: \( a + 3d \) Here, \( a \) is the middle term, and \( d \) is the common difference. ### Step 3: Write the Equation for the Sum of the Terms The sum of these four terms can be expressed as: \[ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 36 \] ### Step 4: Simplify the Equation Now, combine like terms: \[ a - 3d + a - d + a + d + a + 3d = 36 \] This simplifies to: \[ 4a = 36 \] ### Step 5: Solve for \( a \) Now, divide both sides by 4 to find \( a \): \[ a = \frac{36}{4} = 9 \] ### Step 6: Find the Arithmetic Mean (A.M.) The Arithmetic Mean of the four terms is given by the formula: \[ \text{A.M.} = \frac{\text{Sum of the terms}}{\text{Number of terms}} = \frac{(a - 3d) + (a - d) + (a + d) + (a + 3d)}{4} \] From our earlier simplification, we know that: \[ \text{A.M.} = \frac{4a}{4} = a \] Since we found \( a = 9 \): \[ \text{A.M.} = 9 \] ### Final Answer The Arithmetic Mean of the A.P. is **9**. ---
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Knowledge Check

  • If the sum of m terms of an A.P. is same as the sum of its n terms, then the sum of its (m+n) terms is

    A
    mn
    B
    `-mn`
    C
    1/mn
    D
    0
  • The sum of first ten terms of an A.P. is 155 and the sum of first two terms of a G.P. 9. The first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to the common difference of the A.P. which can be the A.P. as per the given conditions?

    A
    2,4,6,8,10,.....
    B
    25/2,79/6,83/6, .....
    C
    2,5,8,11,.....
    D
    both (b) and (c)
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