Question

if the 10th terms of AP is 52 and 17th terms of AP is 20 more than its 13th term​

Answer 1

Answer:

$\huge{\pink{\boxed{\green{\sf{A.P=7,12,17,22,27,32.....}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\to \: a10 = 52 \\ \to \: a + 9d = 52 - - - - - (1) \\ \\ \to \: a17 \: = a13 + 20 \\ \to \: a + 16d = a + 12d + 20 \\ \to \: a - a + 16d - 12d = 20 \\ \to \: 4d = 20 \\ \to \: d = \frac{20}{4} \\ \to \: d = 5$

☆ Putting value of d in (1)

$\to \: a + 9d = 52 \\ \to \: a + 9 \times 5 = 52 \\ \to \: a = 52 - 45 \\ \to \: a = 7$

$\huge{\pink{\boxed{\green{\sf{A.P=7,12,17,22,27,32.....}}}}}$

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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