Question

2.The 7th term of an ap is 32 and its 13 term is 62 find the AP

Answer 1

Answer:

a + 6d=32  - eq 1

a+ 12d=62  - eq 2

by solving both equation we get

d = 5 & a= 2

Then A.P will be as follows:

2,7,12,17,22,27,32,37,42,47,52...................

Step-by-step explanation:

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• 7th term of AP = 32
• 13th term of AP = 62

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• AP = ????

______________________

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\red{\boxed{\bold{t_n = a + ( n - 1)d }}}}$

☆$\sf\implies 32 = a + ( 7 - 1 )d$

$\sf\implies 32 = a + 6d$

$\sf{\bold{\orange{\underline{\underline{ a+ 6d = 32 }}}}}$----- ( i )

☆$\sf\implies 62 = a + ( 13 - 1 )d$

$\sf\implies 62 = a + 12d$

$\sf{\bold{\orange{\underline{\underline{ a+ 12d = 62 }}}}}$----- ( ii )

• Subtracting eq i from ii

$\sf\longrightarrow a + 12d - a - 6d = 62 - 32$

$\sf\longrightarrow 6d = 30$

$\sf\longrightarrow d = \dfrac{30}{6}$

$\sf\longrightarrow d = 5$

$\sf{\blue{\boxed{\bold{d = 5 }}}}$

• putting value of d in eq i

$\sf\longrightarrow a + 6d = 32$

$\sf\longrightarrow a + 6\times 5 = 32$

$\sf\longrightarrow a + 30 = 32$

$\sf\longrightarrow a = 32 - 30$

$\sf\longrightarrow a = 2$

$\sf{\blue{\boxed{\bold{a = 2 }}}}$

• Finding AP

$\sf\leadsto 2 , 2 + 5 , 2 + 10 , 2 + 15 , 2 + 20 , 2 + 25 , 2 + 30 .......$

$\sf\leadsto 2 , 7 , 12 , 17 , 22 , 27 , 32......$

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• AP = 2 , 7 , 12 , 17 , 22 , 27...........