The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is
Step-by-Step Solution
Key Concept: Use the general term of an arithmetic progression $t_n = a + (n-1)d$, where $a$ is the first term and $d$ the common difference. Form equations from the given sums and solve for the required sum.
1. Write the required terms
\[
t_4 = a + 3d, \quad t_8 = a + 7d,\quad t_6 = a + 5d, \quad t_{10}= a + 9d.
\]
2. Use the given information
\[
t_4 + t_8 = (a+3d) + (a+7d) = 2a + 10d = 24. \tag{1}
\]
From (1) we obtain
\[
a + 5d = 12. \tag{2}
\]
(Equation (2) is obtained by dividing (1) by 2.)
3. Express the required sum
\[
t_6 + t_{10} = (a+5d) + (a+9d) = 2a + 14d. \tag{3}
\]
4. Replace $2a$ using (1)
From (1), $2a = 24 - 10d$. Substituting in (3):
\[
t_6 + t_{10} = (24 - 10d) + 14d = 24 + 4d. \tag{4}
\]
5. Conclusion
The sum of the 6th and 10th terms cannot be fixed to a single numeric value unless the common difference $d$ is known. It is given by the expression
\[
\boxed{\;t_6 + t_{10} = 24 + 4d\;}.\]
If additional information (for example, the value of $d$) were supplied, the numerical value could be obtained.
6. Remark (useful for exam)
In many CBSE problems the extra condition is provided (e.g., the sum of the 6th and 10th terms is 28). In that case one would set $24+4d = 28$ to find $d = 1$, and then $a = 12 - 5d = 7$.
However, with the data given in the present statement, the answer remains the expression $24+4d$.
Correct Answer: 24 + 4d (the sum depends on the common difference $d$)