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Question

The $5^{\text{th}}$ and $9^{\text{th}}$ terms of an arithmetic progression are 7 and 13 respectively. What is the $15^{\text{th}}$ term?

The correct answer is
22

Arithmetic Progression Basics

An arithmetic progression (AP) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference ($d$). The formula for the $n^{\text{th}}$ term ($a_n$) is:

$a_n = a + (n-1)d$

where $a$ is the first term.

Using Given Terms to Find $a$ and $d$

We are given:

  • The $5^{\text{th}}$ term ($a_5$) is 7.
  • The $9^{\text{th}}$ term ($a_9$) is 13.

Using the formula $a_n = a + (n-1)d$, we can set up two equations:

  1. For the $5^{\text{th}}$ term: $a_5 = a + (5-1)d \implies a + 4d = 7$
  2. For the $9^{\text{th}}$ term: $a_9 = a + (9-1)d \implies a + 8d = 13$

Subtract equation (1) from equation (2) to eliminate $a$:

$ (a + 8d) - (a + 4d) = 13 - 7 $

$ 4d = 6 $

$ d = \frac{6}{4} = \frac{3}{2} $

Now, substitute the value of $d$ back into equation (1) to find $a$:

$ a + 4\left(\frac{3}{2}\right) = 7 $

$ a + 6 = 7 $

$ a = 7 - 6 = 1 $

Calculating the $15^{\text{th}}$ Term

We need to find the $15^{\text{th}}$ term ($a_{15}$). Using the formula $a_n = a + (n-1)d$ with $n=15$, $a=1$, and $d=\frac{3}{2}$:

$ a_{15} = a + (15-1)d $

$ a_{15} = 1 + (14)\left(\frac{3}{2}\right) $

$ a_{15} = 1 + (7 \times 3) $

$ a_{15} = 1 + 21 $

$ a_{15} = 22 $

The $15^{\text{th}}$ term of the arithmetic progression is 22.

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Important Questions from Progression (Notes)

  1. Find the sum of the G.P.:
    $5/11, 5/121, 5/1331, 5/14641, ...$ to $n$ terms.
  2. What would be the $178^{\text{th}}$ term in the following series?
    2, 5, 8, 11, 14, ...
  3. What is the common difference of the following A.P. series?
    7, 19, 31, 43, ...
  4. The $5^{\text{th}}$ and the $17^{\text{th}}$ terms of an A. P. series are 52 and 184 respectively. Find the common difference of the series.
  5. If a, b and c are in Geometric Progression and $a^\frac{1}{x} = b^\frac{1}{y} = c^\frac{1}{z}$ then, x, y, z are in ________.

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