Question

In an AP a=5, d=3 an=50, Find n and Sn​

Answer 1

This is the solution of ur question.

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Answer 2

$\huge \underline \bold \pink{Solution :}$

As we know that :

$t_{n} = a + (n - 1)d \\ \\ = > 50 = 5 + (n - 1)3 \\ \\ = > 50 - 5 = (n - 1)3 \\ \\ = > 45 = (n - 1)3 \\ \\ = > n - 1 = 15 \\ \\ = >n = 16$

Again we know that :

$S_{n} = \frac{n}{2} [2a + (n - 1)d] \\ \\ = > S_{n} = \frac{16}{2} [2 \times 5 + (16 - 1] \times 3) \\ \\ = > S_{n} = 8(10 +45) \\ \\ = > S_{n} = 8 \times 55 = 440$

$\huge \fbox \green{Hope it helps ☺}$