In an arithmetic progression (A.P.), we can denote the first term by \( a \) and the common difference by \( d \).
The \( n \)-th term of an A.P. can be expressed as:
\[ T_n = a + (n-1)d \]
Given Conditions
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The 3rd term is 10 more than the 1st term: \[ T_3 = a + 2d = a + 10 \] Simplifying this gives: \[ 2d = 10 \ d = 5 \]
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The 5th term is 15 more than the 2nd term: \[ T_5 = a + 4d \quad \text{and} \quad T_2 = a + d \] Therefore, we set up the equation: \[ a + 4d = (a + d) + 15 \] This simplifies to: \[ a + 4d = a + d + 15 \ 4d - d = 15 \ 3d = 15 \ d = 5 \] (Note that we have already determined \( d = 5 \)).
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The 7th term is 7 times the first term: \[ T_7 = a + 6d = 7a \] Substituting \( d = 5 \): \[ a + 6(5) = 7a \ a + 30 = 7a \ 30 = 7a - a \ 30 = 6a \ a = 5 \]
Finding \( a \) and \( d \)
Now we have: \[ a = 5, \quad d = 5 \]
Sum of the 8th and 15th Terms
Now we calculate the 8th term (\( T_8 \)) and the 15th term (\( T_{15} \)): \[ T_8 = a + 7d = 5 + 7(5) = 5 + 35 = 40 \] \[ T_{15} = a + 14d = 5 + 14(5) = 5 + 70 = 75 \]
Sum of the 8th and 15th Terms
Calculating the sum: \[ T_8 + T_{15} = 40 + 75 = 115 \]
Thus, the sum of the 8th and 15th terms of the A.P. is:
\[ \boxed{115} \]