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Find the 10th and 26th terms of the AP: 3, 8, 13, 18, ...
Answer:
First term $a = 3$, common difference $d = 5$.
$10^{th}$ term: $t_{10} = a + 9d = 3 + 9(5) = 3 + 45 = \mathbf{48}$.
$26^{th}$ term: $t_{26} = a + 25d = 3 + 25(5) = 3 + 125 = \mathbf{128}$.
Which term of the AP: 21, 18, 15, ... is -81? Also, is 0 a term of this AP? Give reasons for your answer.
Answer:
First term $a = 21$, common difference $d = -3$.
Let $t_n = -81 \implies 21 + (n-1)(-3) = -81 \implies -3(n-1) = -102 \implies n-1 = 34 \implies n = 35$. So, -81 is the $35^{th}$ term.
Let $t_n = 0 \implies 21 + (n-1)(-3) = 0 \implies -3(n-1) = -21 \implies n-1 = 7 \implies n = 8$. Yes, 0 is a term; it is the $8^{th}$ term because $n$ is a positive integer.
Find the $n^{th}$ term of the AP: 11, 8, 5, 2... Write the recursive rule for this AP.
Answer:
First term $a = 11$, common difference $d = -3$.
$n^{th}$ term (explicit): $t_n = 11 + (n-1)(-3) = \mathbf{14 - 3n}$.
Recursive rule: $\mathbf{t_1 = 11, t_n = t_{n-1} - 3}$ for $n \ge 2$.
An AP consists of 50 terms in which the 3rd term is 12 and the last term is 106. Find the 29th term.
Answer:
$t_3 = a + 2d = 12$
$t_{50} = a + 49d = 106$
Subtracting the first equation from the second gives $47d = 94 \implies d = 2$.
Substitute $d = 2$ back: $a + 4 = 12 \implies a = 8$.
The $29^{th}$ term is $t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = \mathbf{64}$.
How many 2-digit numbers are divisible by 3? What is the sum of all these 2-digit numbers?
Answer:
The 2-digit numbers divisible by 3 form an AP: 12, 15, 18, ..., 99.
Here, $a = 12, d = 3$. Let $t_n = 99 \implies 12 + (n-1)3 = 99 \implies 3(n-1) = 87 \implies n = 30$. There are 30 such numbers.
Sum = $\frac{n}{2}(a + l) = \frac{30}{2}(12 + 99) = 15 \times 111 = \mathbf{1665}$.
Harish started work at an annual salary of ₹5,00,000 and received an increment of ₹20,000 each year. After how many years did his income reach ₹7,00,000?
Answer:
This forms an AP with $a = 500,000$, $d = 20,000$, and $t_n = 700,000$.
$500,000 + (n-1)20,000 = 700,000 \implies 20,000(n-1) = 200,000 \implies n-1 = 10 \implies n = 11$.
His income reached ₹7,00,000 in the $11^{th}$ year, which means it happened after 10 years of receiving increments.
A child arranges marbles in rows so that the first row has 1 marble, the second has 2 marbles, the third has 3, and so on up to 25 rows. How many marbles does the child use in all?
Answer:
The total number of marbles is the sum of the first 25 natural numbers.
$S_{25} = \frac{n(n+1)}{2} = \frac{25 \times 26}{2} = 25 \times 13 = \mathbf{325}$ marbles.
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