# Video: CBSE Class X • Pack 5 • 2014 • Question 16

CBSE Class X • Pack 5 • 2014 • Question 16

04:09

### Video Transcript

If the seventh term of an arithmetic progression is one-ninth and its ninth term is one-seventh, find its 63rd term.

Remember an arithmetic progression is a sequence of numbers such that the difference between each of the consecutive terms is constant. Let’s recall the formula to help us find any term in the sequence. The 𝑛th term is 𝑎 plus 𝑛 minus one multiplied by 𝑑, where 𝑎 is the first term and 𝑑 is the common difference. We’ll first need to calculate the value of the first term and the common difference.

To do so, let’s first substitute what we know about the seventh term into the formula. The seventh term is one-ninth. So that gives us one-ninth is equal to 𝑎 plus seven minus one 𝑑. Seven minus one is six. So we can simplify this expression to give us one-ninth is equal to 𝑎 plus six 𝑑. Similarly, the formula for the ninth term is one-seventh is equal to 𝑎 plus nine minus one multiplied by 𝑑. Nine minus one multiplied by 𝑑 simplifies to eight 𝑑.

Notice that we now have two simultaneous equations. We can solve these to find a value of 𝑎 and 𝑑. First, let’s subtract the equation for the seventh term from the equation we created for the ninth term. This will eliminate the 𝑎 and give us a single equation in terms of 𝑑. 𝑎 minus 𝑎 is zero and eight 𝑑 minus six 𝑑 is two 𝑑. So we get one-seventh minus one-ninth is equal to two 𝑑.

To subtract these fractions, we’ll find the lowest common denominator. The lowest common multiple of seven and nine is 63. So the lowest common denominator of these fractions is 63. To get 63 on the denominator of the first fraction, we multiply by nine. We have to do the same to the numerator. And nine multiplied by one is nine.

Similarly, we multiply the denominator of the second fraction by seven. So we have to do the same to the numerator. One multiplied by seven is seven. So that gives us nine minus seven all over 63 is equal to two 𝑑, which is two over 63 equals two 𝑑. We can solve this equation for 𝑑 by dividing through by two. And we get that 𝑑 is equal to one over 63.

Now that we have the value of 𝑑, we can substitute this back into either of the equations to work out the value of 𝑎. It doesn’t matter which equation we choose. We’ll get the answer either way. Let’s choose the equation for the seventh term. Substituting 𝑑 is equal to one over 63, then gives us one-ninth is equal to 𝑎 plus six multiplied by one over 63.

Six multiplied by one over 63 is six over 63. And we’ll need to subtract this from both sides to calculate the value of 𝑎. 𝑎 is equal to one-ninth minus six over 63. Once again, we choose 63 as the lowest common denominator and we multiply both the numerator and the denominator of our first fraction by seven. That gives us seven minus six all over 63. And we get 𝑎 is one over 63.

So we have both a common difference and a first term of one over 63. To find the 63rd term, we’ll substitute each part into the formula for the 𝑛th term. That gives us one over 63 plus 63 minus one multiplied by one over 63. 63 minus one is 62. So our equation becomes one over 63 plus 62 over 63, which is 63 over 63.

Since 63 over 63 is simply one, the 63rd term in our sequence is one.