Let's solve the problems from the image, starting with question 10.

10. The 8th term of an AP is 31 and its 15th term exceeds its 11th term by 16. Find that AP.

Let the first term of the arithmetic progression (AP) be $a$, and the common difference be $d$.

The general formula for the $n$-th term of an AP is given by:

$T_n = a + (n-1) d$

Given Information:

• The 8th term $T_8 = 31$ a + 7d = 31 \tag{1}
• The difference between the 15th and 11th terms is 16: $T_{15} - T_{11} = 16$ Using the formula for the terms: $(a + 14d) - (a + 10d) = 16$ Simplifying: 4d = 16 \implies d = 4 \tag{2}

Substitute $d = 4$ in equation (1):

$a + 7(4) = 31$ a + 28 = 31 \implies a = 3 \tag{3}

Thus, the first term $a = 3$ and the common difference $d = 4$.

Therefore, the AP is: $3, 7, 11, 15, 19, \dots$

11. How many two-digit positive integers are divisible by 7?

The smallest two-digit number divisible by 7 is $14 \times 7 = 98$, and the largest is $10 \times 7 = 70$.

The two-digit numbers divisible by 7 form an arithmetic progression (AP) with:

• First term $a = 14 \times 7 = 98$
• Common difference $d = 7$
• Last term $l = 70$

The number of terms $n$ in the AP is found using the formula for the last term: $l = a + (n-1) d$ Substituting values: [ 98 = 7 + (10 - 1) d