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Q3(iii):
In the following APs, find the missing terms in the boxes :
(iii) 5, ☐, ☐, $9\frac{1}{2}$
Solution :
Given: An Arithmetic Progression (AP) with the first term $a = 5$ and the fourth term $a_4 = 9\frac{1}{2}$.
To find: The missing terms in the boxes, which correspond to the second term ($a_2$) and the third term ($a_3$).
Step 1: Expressing the given terms using the general formula for an AP.
The general term of an Arithmetic Progression is given by the formula: $a_n = a + (n - 1)d$, where $a$ is the first term, $d$ is the common difference, and $n$ is the position of the term.
Given $a = 5$.
Given $a_4 = 9\frac{1}{2} = \frac{19}{2}$.
Step 2: Formulating the equation for the common difference ($d$).
Using the formula for the fourth term ($n=4$):
$a_4 = a + (4 - 1)d$
$\frac{19}{2} = 5 + 3d$ [Substituting the known values]
Subtract 5 from both sides:
$\frac{19}{2} - 5 = 3d$
$\frac{19 - 10}{2} = 3d$ [Finding a common denominator]
$\frac{9}{2} = 3d$
Divide both sides by 3:
$d = \frac{9}{2 \times 3}$
$d = \frac{3}{2}$ or $1.5$
Step 3: Calculating the missing terms.
Now that we have $a = 5$ and $d = \frac{3}{2}$, we can find the missing terms $a_2$ and $a_3$.
For the second term ($a_2$):
$a_2 = a + d$
$a_2 = 5 + \frac{3}{2}$
$a_2 = \frac{10 + 3}{2} = \frac{13}{2} = 6\frac{1}{2}$
For the third term ($a_3$):
$a_3 = a + 2d$
$a_3 = 5 + 2(\frac{3}{2})$
$a_3 = 5 + 3 = 8$
Verification:
The sequence is $5, 6\frac{1}{2}, 8, 9\frac{1}{2}$.
Common difference check: $6.5 - 5 = 1.5$; $8 - 6.5 = 1.5$; $9.5 - 8 = 1.5$. Since the common difference is constant, the values are correct.
Final Answer: The missing terms are $6\frac{1}{2}$ and $8$.
More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.2
- Q1(i): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (i) $a=7, d=3, n=8, a_n=...$
- Q1(ii): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (ii) $a=–18, d=..., n=10, a_n=0$
- Q1(iii): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (iii) $a=..., d=–3, n=18, a_n=–5$
- Q1(iv): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (iv) $a=–18.9, d=2.5, n=..., a_n=3.6$
- Q1(v): Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n$th term of the AP: (v) $a=3.5, d=0, n=105, a_n=...$
- Q10: The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
- Q11: Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term?
- Q12: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
- Q13: How many three-digit numbers are divisible by 7?
- Q14: How many multiples of 4 lie between 10 and 250?
- Q15: For what value of $n$, are the $n$th terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal?
- Q16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.
- Q17: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253.
- Q18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
- Q19: Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?
- Q2(i): Choose the correct choice in the following and justify : (i) 30th term of the AP: 10, 7, 4, . . . , is
- Q2(ii): Choose the correct choice in the following and justify : (ii) 11th term of the AP: – 3, $-\frac{1}{2}$, 2, . . ., is
- Q20: Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$th week, her weekly savings become ₹ 20.75, find $n$.
- Q3(i): In the following APs, find the missing terms in the boxes : (i) 2, ☐, 26
- Q3(ii): In the following APs, find the missing terms in the boxes : (ii) ☐, 13, ☐, 3
- Q3(iv): In the following APs, find the missing terms in the boxes : (iv) – 4, ☐, ☐, ☐, ☐, 6
- Q3(v): In the following APs, find the missing terms in the boxes : (v) ☐, 38, ☐, ☐, ☐, – 22
- Q4: Which term of the AP : 3, 8, 13, 18, . . . ,is 78?
- Q5(i): Find the number of terms in each of the following APs : (i) 7, 13, 19, . . . , 205
- Q5(ii): Find the number of terms in each of the following APs : (ii) 18, $15\frac{1}{2}$, 13, . . . , – 47
- Q6: Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . .
- Q7: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
- Q8: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
- Q9: If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
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