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Question
The 26th, 11th and the last term of an AP are 0, 3 and `- 1/5`, respectively. Find the common difference and the number of terms.
Solution
Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
We know that, if last term of an AP is known, then
l = a + (n – 1)d ...(i)
And nth term of an AP is
Tn = a + (n – 1)d ...(ii)
Given that,
26th term of an AP = 0
⇒ T26 = a + (26 – 1)d = 0 ...[From equation (i)]
⇒ a + 25d = 0 ...(iii)
11th term of an AP = 3
⇒ T11 = a + (11 – 1)d = 3 ...[From equation (ii)]
⇒ a + 10d = 3 ...(iv)
And last term of an AP = `-1/5`
⇒ l = a + (n – 1)d ...[From equation (i)]
⇒ `-1/5` = a + (n – 1)d ...(v)
Now, subtracting equation (iv) from equation (iii),
a + 25d = 0
a + 10d = 3
– – –
15d = – 3
⇒ d = `- 1/5`
Put the value of d in equation (iii), we get
`a + 25(-1/5)` = 0
⇒ a – 5 = 0
⇒ a = 5
Now, put the value of a, d in equation (v), we get
`-1/5 = 5 + (n - 1)(-1/5)`
⇒ – 1 = 25 – (n – 1)
⇒ – 1 = 25 – n + 1
⇒ n = 25 + 2 = 27
Hence, the common difference and number of terms are `-1/5` and 27, respectively.
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