Important Questions: Arithmetic Progressions - Class 10 MCQ

∵ p, q, r are in AP.
∴ 2 q = p + r
⇒ p + r - 2 q = 0

Let next term be T4

The given AP is √8, √18, √32,......... On simplifying the terms, we get: 

Johann Friedrich Gauss, he was a German mathematician who find the sum of the first 100 natural number.

The given AP is 5, 2, -1, -4 ... ...., -31
d = 2 - 5 = -3, so d for the AP starting from the last term is 3.
The first term = l = -31
We know, a_n = a+(n−1)d
a₆ from the end = −31+(5)3
a₆ from the end = −16

a₄ - b₄ = (a₁ + 3d) - (b₁ + 3d)
= a₁ b₁ = - 1 - (-8) = 7

Given:
S_n​ = 3n² + 2n
For an AP, the nth term is
a_{n ​}= S_n​ − S_{n − 1}
Step 1: Find S_n−1
S_n−1 = 3(n−1)² + 2(n−1)
= 3(n²−2n+1) + 2n−2
= 3n² − 6n + 3 + 2n − 2
= 3n² − 4n + 1
Step 2: Find a_n
a_n ​= S_n​ − S_n−1​
= (3n² + 2n) − (3n² − 4n + 1)
= 6n − 1
So, a_n = 6n − 1
We can calculate:
a₁ = 6 × 1 - 1 = 6 - 1 = 5
a₂ = 6  × 2 - 1 = 12 - 1 = 11
a₃ = 6  × 3 - 1 = 18 - 1 = 17 
......and so on we can find a₄, a₅ ,....
Required A.P = 5, 11, 17, ......
Common Difference (d) = a₂ - a₁ = 11 - 5 = 6

So, Correct Answer: 6

Let nth term of the given AP be 210.
Here, first term,
a = 21
And common difference,
d = 42 – 21 = 21 and
a_n = 210
a_n = a + (n-1)d
⇒ 210 = 21 + (n - 1)21
⇒ 210 = 21 + 21n - 21
⇒ 210 = 21n
⇒ n = 10
Hence, the 10th term of the AP is 210.

The first term aaa is positive.
For the sum of the first nnn terms to become zero, some later terms must be negative so that the positive terms get cancelled.

This is possible only when the common difference d is negative, because a negative ddd makes the AP decrease and eventually the terms become negative.

Hence,

S_n = 0 is possible only when d < 0.

To find the total savings of the boy over 12 months, we first observe the pattern in his savings: he saves ₹50 in the first month, ₹60 in the second month, ₹70 in the third month, and so on. This shows that he increases his savings by ₹10 each month.

The savings for each month can be written as:

• Month 1: ₹50
• Month 2: ₹60
• Month 3: ₹70
• Month 4: ₹80
• Month 5: ₹90
• Month 6: ₹100
• Month 7: ₹110
• Month 8: ₹120
• Month 9: ₹130
• Month 10: ₹140
• Month 11: ₹150
• Month 12: ₹160

Now, we can calculate the total savings by adding the amounts saved each month. The total can be calculated using the formula for the sum of an arithmetic series:

Total Savings = Number of Months × (First Month Saving + Last Month Saving) / 2

Here, the number of months is 12, the first month saving is ₹50, and the last month saving is ₹160.

Total Savings = 12 × (₹50 + ₹160) / 2

Total Savings = 12 × ₹210 / 2

Total Savings = 12 × ₹105

Total Savings = ₹1260

Thus, the boy's total savings over 12 months are ₹1260.