Question

Example of an AP in series​

Answer 1

Answer:

$\huge\blue{HELLO\:MATE}$

IN AN AP

LET a BE THE 1ST TERM

d BE THE COMMON DIFFERENCE

AND

n BE THE NO. OF TERMS

THEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]

IN A GP

LET a BE THE 1ST TERM

r BE THE CONSTANT RATIO

AND

n BE THE NO. OF TERMS

THEN SUM OF N TERMS (Sn)=

$a \dfrac{( {r}^{n} - 1) }{(r - 1)}$

Answer 2

Explanation:

➪@ɴsᴡᴇʀ;-

IN AN AP

IN AN APLET a BE THE 1ST TERM

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCE

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEAND

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMS

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]IN A GP

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]IN A GPLET a BE THE 1ST TERM

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]IN A GPLET a BE THE 1ST TERMr BE THE CONSTANT RATIO

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]IN A GPLET a BE THE 1ST TERMr BE THE CONSTANT RATIOAND

IN AN APLET a BE THE 1ST TERMd BE THE COMMON DIFFERENCEANDn BE THE NO. OF TERMSTHEN SUM OF N TERMS (Sn)=n/2[2a+(n-1)d]IN A GPLET a BE THE 1ST TERMr BE THE CONSTANT RATIOANDn BE THE NO. OF TERMS (NP)

$a(r−1)(rn−1)</strong></p><p><strong>[tex]a(r−1)(rn−1)$