Question

the 10th term of an AP is 52 and its 17th term exceeds its 13th term by 20 find the AP and also its 30th term

Answer 1

In Arithmetic Progressions,  

a = first term

d = common difference

a₂ = second term = a + d  

aₓ = x th term

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Given that 0th term of an AP is 52 and its 17th term exceeds its 13th term by 20.

Now,

10th term = a + 9d = 52

⇒ a + 9d = 52

⇒ a = 52 - 9d         ...( i )

17th exceeds 13th term by 20.

∴ 17th term = 13th + 20

⇒ a + 16d = ( a + 12d ) + 20

⇒ a + 16d = a + 12d + 20

⇒ a - a + 16d - 12d = 20

⇒ 4d = 20

⇒ d = $\dfrac{20}{4}$

⇒ d = 5

Substituting the value of d in ( i )

⇒ a = 52 - 9( 5 )

⇒ a = 52 - 45

⇒ a = 7

∴ 30 th term = a + ( 30 - 1 )d

                     = 7 + 29( 5 )

                     = 7 + 145

                     = 152

Then, as a is 7, first term of the AP is 7.

∴ Required AP : a , a + d , a + 2d , a + 3d , a + 4d .....

                         : 7 , 7 + 5 , 7 + 2( 5 ) , 7 + 3( 5 ) , 7 + 4( 5 ) .....

                         : 7 , 12 , 7 + 10 , 7 + 15 , 7 + 20 ......

                         : 7 , 12 , 17 , 22 , 27 .....

therefore required AP is 7 , 12 , 17 , 22 , 27 .... and 30th term is 152.

Answer 2

Answer:

Step-by-step explanation:

Given the 10'th term of an AP is 52 and its 17'th term is 20 more than its 13'th term

We have to find out,

The AP and its 30'th term

10'th term is 52 i.e. t₁₀ = a + 9d

a + 9d = 52 (given) -----[1]

its 17th term is 20 more than its 13th term

t₁₇ = 20 + t₁₃

⇒ a + 16d = 20 + a + 12d

⇒ 4d - 20 = 0

⇒ 4d = 20

⇒ d = 5

Substitute 'd' in [1]

We get,

⇒ a + 9d = 52

⇒ a + 45 = 52

⇒ a = 7

30'th term = a + 29d

7 + (29 * 5) = 152

Required answer:

AP = 7 , 12 , 17 , 22 , 27

30'th term = 152