Question

$\mathfrak{question}$

if the sum of first 14 terms pf an arithmetic progression is 1050 and its fourth term is 40, find its 20th term.


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Answer 1

$\dag\:\underline{\sf AnsWer :}$

$\frak {\pink{Given}}\begin{cases} \tt{\green{Fourth \: term \: of \: a \: A.P \:( a_{4}) = 40}}\\ \\ \tt{\blue{20^{th} \: term \: of \: a \: A.P \: ( a_{20}) = \: ?}} \\ \\ \sf{\orange{Sum \: of \: first \: 14 \: terms \: S_{14} = 1050}}\end{cases}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies \sf a_n= a + (n - 1)d$

$:\implies \sf a_4= a + (4 - 1)d$

$:\implies \sf 40= a + 3d \qquad...(i)$

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$\dashrightarrow\:\:\sf S = \dfrac{n}{2} \bigg \lgroup 2a + (n - 1)d \bigg \rgroup$

$\dashrightarrow\:\:\sf S_{14} = \dfrac{14}{2} \bigg \lgroup 2a + (14 - 1)d \bigg \rgroup$

$\dashrightarrow\:\:\sf 1050 = \dfrac{14}{2} \bigg \lgroup 2a + (14 - 1)d \bigg \rgroup$

$\dashrightarrow\:\:\sf 1050 = 7 \bigg \lgroup 2a + 13d \bigg \rgroup$

$\dashrightarrow\:\:\sf \dfrac{1050}{7} = 2a + 13d$

$\dashrightarrow\:\:\sf 150 = 2a + 13d$

$\dashrightarrow\:\:\sf 75 = a + 6.5d \quad...(ii) \qquad \bigg \lgroup \bf{Dividing \: whole \: {eq}^{n} \: by \: 2} \bigg \rgroup$

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$\tiny\underline{\frak{Now, from\: \: eq^n \: (i) \: and \: eq^n \: (ii) \: we \: get :}}$

$\longrightarrow\:\:\sf 75 - 40 = a + 6.5 d - (a + 3d)$

$\longrightarrow\:\:\sf 35 = a + 6.5 d - a - 3d$

$\longrightarrow\:\:\sf 35 = 6.5 d - 3d$

$\longrightarrow\:\:\sf 35 = 3.5d$

$\longrightarrow\:\:\sf d = \dfrac{35}{3.5}$

$\longrightarrow\:\: \underline{ \boxed{\sf d = 10 }}$

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$\tiny\underline{\frak{Substituting \: the \: value \: of \: d = 10 \: in \: eq^n \: (i) \: we \: get :}}$

$\leadsto\sf 40= a + 3d$

$\leadsto\sf 40= a + 3 \times 10$

$\leadsto\sf 40= a + 30$

$\leadsto\sf a = 40 - 30$

$\leadsto \underline{ \boxed{\sf a = 10}}$

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$\qquad\dag\:\underline{\sf 20^{th} \: term \: of \: a \: A.P}$

$\mapsto\:\:\sf a_{n} = a + (n - 1)d$

$\mapsto\:\:\sf a_{20} = 10 + (20 - 1) \times 10$

$\mapsto\:\:\sf a_{20} = 10 + 19 \times 10$

$\mapsto\:\:\sf a_{20} = 10 + 190$

$\mapsto\:\: \underline{ \boxed{\sf a_{20} =200}}$

Answer 2

Answer:

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Step-by-step explanation:

S 14 =1050

a=10

Sn = n/2[2a+(n−1)d]

1050 = 14/2(2×10+(14−1)d])

1050/7 = 20+14d−d]

150−20=13d

130=13d

d=10

The 20 th term is

a20 =a+(n−1)d

=10+(20−1)10

=10+190

a20=200

Hope it is helpful