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The 7th Term of an A.P. is 32 and Its 13th Term is 62. Find the A.P. - Mathematics

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प्रश्न

The 7th term of an A.P. is 32 and its 13th term is 62.  Find the A.P.

योग

उत्तर

Here, let us take the first term of the A.P. as a and the common difference of the A.P as d

Now, as we know,

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

So for the 7th term (n = 7)

`a_7 = a + (7 - 1)d`

32 = a + 6d .......(1)

Also for 12th term (n = 13

`a_13 = a + (13 - 1)d`

62 = a + 12d ......(2)

Now on substracting (2) from (1) we get

62 - 32 = (a + 12d) - (a + 6d)

30 = a + 12d - a - 6d

30 = 6d

`d = 30/6`

d = 5

Substituting the value of d in (1) we get

32 = a + 6(5)

32 = a + 30

a = 32 - 30

a = 2

So, the first term is 2 and the common difference is 5.

There the A.P is 2, 7, 12, 17, ....

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अध्याय 5: Arithmetic Progression - Exercise 5.4 [पृष्ठ २५]

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आरडी शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progression
Exercise 5.4 | Q 25 | पृष्ठ २५
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