Arithmetic Progressions - Free MCQ Practice Test with solutions,

To find the 18^th term of the A.P., we need to determine its first term and common difference.

Given:

• 7^th term (T₇) = 34
• 13^th term (T₁₃) = 64

Formulas:

• T_n = a + (n - 1)d

Using the given terms:

• T₇ = a + 6d = 34
• T₁₃ = a + 12d = 64

We can create a system of equations:

• Equation 1: a + 6d = 34
• Equation 2: a + 12d = 64

Subtract Equation 1 from Equation 2:

• (a + 12d) - (a + 6d) = 64 - 34
• 6d = 30
• d = 5

Now substitute d back into Equation 1:

• a + 6(5) = 34
• a + 30 = 34
• a = 4

Now we can find the 18^th term:

• T₁₈ = a + 17d
• T₁₈ = 4 + 17(5)
• T₁₈ = 4 + 85 = 89

The 18^th term is 89.

Here a = 11 and d = 2, t_n= 99, n = ?
Sum of the n terms = (n/2)[2a+(n -1)d]
But t_n = a + (n -1)d
⇒ 99 = 11+ (n-1)2
⇒ 99 -11 = (n-1)2
⇒ 88/2 = (n-1)
∴ n = 45.
subsitute n = 45  in sum of the n terms we obtain
⇒ s_{45 =} (45/2)(2×11 + (45 -1)2)
⇒ s_{45 =} (45/2)(110)
⇒ s₄₅ = 45×55.
⇒  s₄₅ = 2475.
∴ sum of all two digit odd positive numbers = 2475.

example :
 

Given:

The nth term of the A.P. is aₙ = 3n + 4.

Number of terms n = 12.

Step 1: Find the first term a₁
a₁ = 3(1) + 4 = 3 + 4 = 7

Step 2: Find the 12th term a₁₂
a₁₂ = 3(12) + 4 = 36 + 4 = 40

Step 3: Use the sum formula for the first n terms of an A.P.:
Sₙ = (n/2)(a₁ + aₙ)

For n = 12:
S₁₂ = (12/2)(7 + 40) = 6 × 47 = 282

Answer:
The sum of the first 12 terms is 282.

Explanation: For an AP,

a_n = a+(n-1)d

= 28+(7-1)(-4)

= 28+6(-4)

= 28-24

a_n=4