If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term

Solution:

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.

Consider the first term, common difference and number of terms of an AP are a, d and n.

We know that,

9th term of an AP, T₉ = 0

So the nth term of an AP, Tn = a + (n - 1 )d

⇒ a + (9 - 1)d = 0

⇒ a + 8d = 0 ⇒ a = -8 d ………….. (i)

The 19th term, T₁₉ = a + (19 - 1)d

= -8d + 18d [From Equation (i)]

= 10d …………… (ii)

The 29th term, T₂₉ = a + (29 - 1)d

= -8d + 28d

= 20d

= 2 × (10d) [from Equation (i)]

⇒ T₂₉ = 2 × T₁₉

Therefore, it is proved.

✦ Try This: If the 10th term of an AP is zero, prove that its 30th term is twice its 20th term

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.3 Problem 9

If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term

Summary:

If the 9th term of an AP is zero, it is proved that its 29th term is twice its 19th term

☛ Related Questions:

• Determine k so that k²+ 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP
• Split 207 into three parts such that these are in AP and the product of the two smaller parts is 462 . . . .
• The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of th . . . .