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The 26th, 11th and the last term of an AP are 0, 3 and -15, respectively. Find the common difference and the number of terms. - Mathematics

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प्रश्न

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.

योग

उत्तर

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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अध्याय 5: Arithematic Progressions - Exercise 5.3 [पृष्ठ ५२]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
अध्याय 5 Arithematic Progressions
Exercise 5.3 | Q 6 | पृष्ठ ५२
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