Question

If the fifth term of a AP is 19 and the tenth term is 34
Find:- a) the first term
b) The common difference
c) The sum of the first 15 term

Answer 1

Answer:

Here's your answer

➡a+4d =19

➡a +9d =34

----------------

-5d =-15

➡ d =3

⏩Then a will be...

➡a +4d =19

➡a =19 -12

➡a =7

Sum =?

$sum = \frac{15}{2} (2 \times 7 + 14 \times 3) \\ \\ = \frac{15}{2} \times 56 \\ \\ = 15 \times 28 \\ \\ = 420$

Hope it helps ❗

Answer 2

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Here is ur sweet answer

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$\:\:\:\: \: \: \: \: \: \: \: \: \: \: \large\mathfrak{\underline{\underline{\huge\mathcal{\bf{\boxed{\huge\mathcal{~~QUESTION~~}}}}}}}$

If the fifth term of a AP is 19 and the tenth term is 34

Find:- a) the first term

b) The common difference

c) The sum of the first 15 term

______________________________________________

$\:\: \underline{\underline{\bf{\large\mathfrak{~~solution~~}}}}$

let the first term of the AP series is a and the common difference of the AP series is d respectively.

now it is given that the fifth term is 19...

so according to the problem

$=>19=a+(5-1)d\\=>19=a+4d\\=>a=19-4d$

now it is also given that the 10th term is 34

again applying the formula of AP series we get.

$=>34=a+(10-1)d\\=>34-9d=a\\=>34-9d=19-4d\\=>5d=15\\=>d=3\\</p><p>therefore.....a=19-4(3)=7$

therefore......

a)

the first term is 7

b)

the common difference is 3

C)

the sum of the first 15 terms is...

$=\frac{15(2\times7+(15-1)3)}{2}\\=15\times28\\=420$

$\huge\mathcal\green{\underline{hope\:\: this\:\: helps\:\: you}}$