Question

the 10th ,28th and last term of an AP are respectively 29,83 and 122 . find the first term and the common difference and the number of terms of the AP​

Answer 1

Answer:

• T₁₀ = 29
• T₂₈ = 83
• Last Term = 122

☯ According to the Question :

⇴ T₁₀ = 29

⇴ a + (10 – 1)d = 29

⇴ a + 9d = 29

⇴ a = 29 – 9d ⠀— eq. ( I )

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⇴ T₂₈ = 83

⇴ a + (28 – 1)d = 83

⇴ a + 27d = 83

• Putting value of a from eq. ( I )

⇴ 29 – 9d + 27d = 83

⇴ 18d = 83 – 29

⇴ 18d = 54

• Dividing both term by 18

⇴ d = 3

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☯ Putting value of d in eq. ( I ) :

⇾ a = 29 – 9d

⇾ a = 29 – 9(3)

⇾ a = 29 – 27

⇾ a = 2

∴ Hence, First Term and Common Difference of AP is 2 and 3 respectively.

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☢ Last Term of the AP :

⇢ Tn = a + (n – 1)d

⇢ 122 = 2 + (n – 1)3

⇢ 122 – 2 = (n – 1)3

⇢ 120 = (n – 1)3

• Dividing both term by 3

⇢ 40 = n – 1

⇢ 40 + 1 = n

⇢ n = 41

∴ Hence, Number of Terms in AP is 41.

Answer 2

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$\huge \tt {GIVEN:}$

• 10th term = 29
• 28th term = 83
• Last term = 122

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$\huge \tt {SOLUTION:}$

10th term = 29

↪a + (10-1)D = 29

↪a + 9D = 29

↪a = 29 - 9D______(EQ.1)

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↪28th term = 83

↪a +(18-1)D = 83

↪a + 27D = 83

Comparing value of (EQ.1),

↪29-9D + 27d = 83

↪18D = 83-29

↪18D = 54

↪D = 54/18

↪D = 3

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Now,

Again Comparing values of (EQ.1)

↪a = 29 -9D

↪a = 29 - 9×3

↪a = 29-27

↪a = 2

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Therefore,

↪First term = 2

↪Common differences = 3

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Now,

the last term-

↪Nth term = a + (n - 1)d

↪122 = 2 + (n-1)3

↪122 -2 = (n-1)3

↪120 = (n-1)3

↪40 = n-1

↪40+1 = n

↪N = 41

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