Question

If sums of first 8 and 19 terms of


a.p. are 64 and 361 find common difference and sum of n terms of series

Answer 1

Answer:

$\green { a = 1, \: d= 2,\:S_{n} = n^{2}}$

Step-by-step explanation:

$Let \: a \: and \: d \: are \: first \: term \: and \\common\: difference \: of \: an \: A.P$

$\boxed {\blue { Sum \: of \: first \:n \:terms (S_{n}) = \frac{n}{2}[2a+(n-1)d]}}$

$i)Sum \: of \: 8\:terms (S_{8}}= 64$

$\frac{8}{2}(2a+7d)=64$

$\implies 4(2a+7d) =64$

$\implies 2a+7d = \frac{64}{4}$

$\implies 2a+7d = 16\:---(1)$

$ii)Sum \: of \: 19\:terms (S_{19})= 361$

$\frac{19}{2}(2a+18d)=361$

$\implies 2a+18d = \frac{361\times 2}{19}$

$\implies 2a+18d = 38\:---(2)$

/* Subtract equation (1) from equation (2) ,we get

$\implies 11d = 22$

$\implies d = 2$

/* Substitute d = 2 in equation (1) ,we get

$2a + 7\times 2 = 16$

$\implies 2a + 14 = 16$

$\implies 2a = 2$

$\implies a = \frac{2}{2} = 1$

$\blue { a = 1 , \:d = 2 } , \red { S_{n} = ? }$

$S_{n} = \frac{n}{2}[2\times 1 + (n-1)2]$

$= \frac{n}{2}\times 2[1+n-1]$

$= n\times n = \green {n^{2}}$

Therefore.,

$\green { a = 1, \: d= 2,\:S_{n} = n^{2}}$

•••♪

Answer 2

Answer:

• Sum of n terms in AP :

Sn = (n/2)[2a + (n- 1)d]

───────────────

⇒ S₈ = 64

⇒ 8/2 × (2a + 7d) = 64

⇒ 4 × (2a + 7d) = 64

⇒ 2a + 7d = 16 — eq. ( I )

⇒ S₁₉ = 361

⇒ 19/2 × (2a + 18d) = 361

⇒ 19 × (a + 9d) = 361

⇒ a + 9d = 19 — eq. ( II )

• Multiplying eq.( II ) by 2 & Subtracting from eq.( I ) from eq.( II ) :

↠ 2a + 18d - 2a - 7d = 38 - 16

↠ 11d = 22

↠ d = 2

• Substitute d value in eq. ( II ) :

⇒ a + 18 = 19

⇒ a = 19 - 18

⇒ a = 1

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⋆ Sum of nth terms of the AP :

↠ Sn = n/2 [2a + (n - 1)d]

↠ Sn = n/2 × [2 × 1 + (n - 1) × 2]

↠ Sn = n/2 × [2 + 2n - 2]

↠ Sn = n/2 × 2n

↠ Sn = n × n

↠ Sn = n²

∴ Sum of nth terms of the AP is n².