Question

In an ap 19 term is 52 and 38th term is 128 find the sum of first ppt

Answer 1

$\huge \bold{AP}$

AP is abbreviation for Arithmetic progression, its a series in which numers occurs after a fixed interval of gap. The difference between the next term and previous term is always constant and is known as common difference.

FOR AN AP

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

where a is first term, d is common difference and n is number of terms.

ACCORDING TO QUESTION :--

19th term is equal to 52.

$\mathsf {t_19 = a + (19 - 1) d }$

$\mathsf {52 = a + (19 - 1) d }$

$\mathsf {52 = a + 18d }$----(1)

38th term is equal to 128.

$\mathsf {t_38 = a + (38 - 1) d }$

$\mathsf {128= a + (19 - 1) d }$

$\mathsf {128= a + 37d }$----(1)

On subtracting both equations

$\mathsf {128= a + 37d }$

$\mathsf {52 = a + 18d }$

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76 = 19d

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$\mathsf{\dfrac{76}{19} = d}$

$\boxed{ \mathsf{4 = d}}$

On placing value in equation (1)

$\mathsf {52 = a + 18d }$

$\mathsf {52 = a + 18 \times 4}$

$\mathsf {52 = a + 72}$

$\mathsf {52 - 72= a }$

$\boxed{ \mathsf{-10 = a}}$

FOR AN AP SUM OF n terms is

$\boxed{ \mathsf{ S_n = \frac{n}{2} (2a + (n-1)d)}}$

$\mathsf{ S_n = \frac{n}{2} (2 \times (-10)+ (n-1) \times 4)}$

$\mathsf{ S_n = \frac{n}{2} ( -20+ 4n - 4)}$

$\mathsf{ S_n = \frac{n}{2} ( -24+ 4n)}$